6 = 1 2. The right endpoints of the subintervals are then 2 5, 3, 7 2, 4, 2 9, 5, while the left endpoints are 2, 5 2, 3, 7 2, 4, 9 2.

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1 5 THE ITEGRAL 5. Approimating and Computing Area Preliminar Questions. What are the right and left endpoints if [, 5] is divided into si subintervals? If the interval [, 5] is divided into si subintervals, the length of each subinterval is 5 6 =. The right endpoints of the subintervals are then 5,, 7,, 9, 5, while the left endpoints are, 5,, 7,, 9.. The interval [, 5] is divided into eight subintervals. a What is the left endpoint of the last subinterval? b What are the right endpoints of the first two subintervals? ote that each of the 8 subintervals has length 5 8 =. a The left endpoint of the last subinterval is 5 = 9. b The right endpoints of the first two subintervals are + = and + =.. Which of the following pairs of sums are not equal? a i, l b i= l= 5 c j, i d j= i= 5 j, k j= k= ii +, i= 5 j j j= a Onl the name of the inde variable has been changed, so these two sums are the same. b These two sums are not the same; the second squares the numbers two through five while the first squares the numbers one through four. c These two sums are the same. ote that when i ranges from two through five, the epression i ranges from one through four. d These two sums are the same. Both sums are Eplain: j= j = j= j but j= is not equal to. j= The first term in the sum j= j is equal to zero, so it ma be dropped. More specificall, j = + j = j. j= On the other hand, the first term in j= is not zero, so this term cannot be dropped. In particular, j= j= = +. j= j= j= 5. Eplain wh L R for f= on [, 7]. On [, 7], the function f= is a decreasing function; hence, for an subinterval of [, 7], the function value at the left endpoint is larger than the function value at the right endpoint. Consequentl, L must be larger than R. 565 April,

2 566 CHAPTER 5 THE ITEGRAL Eercises. Figure 5 shows the velocit of an object over a -min interval. Determine the distance traveled over the intervals [, ] and [,.5] remember to convert from km/h to km/min. km/h FIGURE 5 min The distance traveled b the object can be determined b calculating the area underneath the velocit graph over the specified interval. During the interval [, ], the object travels =.96 km During the interval [,.5], it travels = =.5 km An ostrich Figure 6 runs with velocit km/h for minutes, km/h for minutes, and km/h for another minute. Compute the total distance traveled and indicate with a graph how this quantit can be interpreted as an area. FIGURE 6 Ostriches can reach speeds as high as 7 km/h. The total distance traveled b the ostrich is + + = = 9 5 km. This distance is the area under the graph below which shows the ostrich s velocit as a function of time A rainstorm hit Portland, Maine, in October 996, resulting in record rainfall. The rainfall rate Rt on October is recorded, in centimeters per hour, in the following table, where t is the number of hours since midnight. Compute the total rainfall during this -hour period and indicate on a graph how this quantit can be interpreted as an area. t h 9 9 Rt cm Thus Over each interval, the total rainfall is the time interval in hours times the rainfall in centimeters per hour. R = = 8.5 cm. The figure below is a graph of the rainfall as a function of time. The area of the shaded region represents the total rainfall. April,

3 SECTIO 5. Approimating and Computing Area The velocit of an object is vt = t m/s. Use Eq. and geometr to find the distance traveled over the time intervals [, ] and [, 5]. B equation Eq., the distance traveled over the time interval [a,b] is b b vt dt = t dt; a a that is, the distance traveled is the area under the graph of the velocit function over the interval [a,b]. The graph below shows the area under the velocit function vt = t m/s over the intervals [, ] and [, 5]. Over the interval [, ], the area is a triangle of base and height ; therefore, the distance traveled is = meters. Over the interval [, 5], the area is a trapezoid of height and base lengths and 6; therefore, the distance traveled is + 6 = 6 meters Compute R 5 and L 5 over [, ] using the following values f = 5 =.. Thus, L 5 = =. = 6, and R 5 = =. =. The average is 6 + = 5. This estimate is frequentl referred to as the Trapezoidal Approimation. 6. Compute R 6, L 6, and M to estimate the distance traveled over [, ] if the velocit at half-second intervals is as follows: t s v m/s 8 5 For R 6 and L 6, t = 6 =.5. For M, t = =. Then R 6 =.5 s m/s =.59 m = 5.5m, April,

4 568 CHAPTER 5 THE ITEGRAL L 6 =.5 sec m/sec =.589 m =.5m, and M = sec m/sec = 5 m. 7. Let f= +. a Compute R 6 and L 6 over [, ]. b Use geometr to find the eact area A and compute the errors A R 6 and A L 6 in the approimations. Let f = + on[, ]. a We partition [, ] into 6 equall-spaced subintervals. The left endpoints of the subintervals are { } whereas the right endpoints are,,,, 5,. { },,,,, 5 Let a =, b =, n = 6, = b a /n =, and k = a + k, k =,,...,5 left endpoints. Then L 6 = 5 f k = k= With k = a + k, k =,,...,6 right endpoints, we have R 6 = 6 f k = k= 5 f k = = 6.5. k= 6 f k = = 9.5. k= b Via geometr see figure below, the eact area is A = 6 + = 8. Thus, L 6 underestimates the true area L 6 A =.5, while R 6 overestimates the true area R 6 A = Repeat Eercise 7 for f = over [, ]. Let f= on [, ]. a We partition [, ] into 6 equall-spaced subintervals. The left endpoints of the subintervals are { whereas the right endpoints are 7, 8,,, }., { }, 7, 8,,, Let a =, b =, n = 6, = b a /n =, and k = a + k, k =,,...,5 left endpoints. Then L 6 = 5 f k = k= With k = a + k, k =,,...,6 right endpoints, we have R 6 = 6 f k = k= 5 f k = =. k= 6 f k = =. k= b Via geometr see figure below, the eact area is A = + 8 =. Thus, L 6 overestimates the true area L 6 A =, while R 6 underestimates the true area R 6 A =. 8 6 April,

5 SECTIO 5. Approimating and Computing Area Calculate R and L for f= + over [, ] Then sketch the graph of f and the rectangles that make up each approimation. Is the area under the graph larger or smaller than R? Is it larger or smaller than L? Let f= + and set a =, b =, n =, = b a /n = / =. a Let k = a + k, k =,,,. Selecting the left endpoints of the subintervals, k, k =,,, or {,, }, we have L = f k = k= f k = =. Selecting the right endpoints of the subintervals, k, k =,,, or {,, }, we have R = f k = k= k= f k = =. k= b Here are figures of the three rectangles that approimate the area under the curve fover the interval [, ]. Clearl, the area under the graph is larger than L but smaller than R. L R Let f = + and =. Sketch the graph of fand draw the right-endpoint rectangles whose area is 6 represented b the sum f + i. i= Because = and the sum evaluates f at + i for i from through 6, it follows that the interval over which we are considering f is [, ]. The sketch of f together with the si rectangles is shown below Estimate R, M, and L 6 over [,.5] for the function in Figure FIGURE 7 } Let f on [, ] be given b Figure 7. For n =, = / =, { k} k= {, =,,. Therefore R = f k = + + =.5, k= April,

6 57 CHAPTER 5 THE ITEGRAL M = 6 f k= For n = 6, = /6 =, { k} 6 k= = {,,,,, 5, k = =.875. }. Therefore L 6 = 5 f k = =.75. k=. Calculate the area of the shaded rectangles in Figure 8. Which approimation do these rectangles represent? = + FIGURE 8 Each rectangle in Figure 8 has a width of and the height is taken as the value of the function at the midpoint of the interval. Thus, the area of the shaded rectangles is = Because there are si rectangles and the height of each rectangle is taken as the value of the function at the midpoint of the interval, the shaded rectangles represent the approimation M 6 to the area under the curve. In Eercises, calculate the approimation for the given function and interval.. R, f= 7, [, 5] { Let f= 7 on [, 5]. For n =, = 5 / =, and { k} k= =,, }., 5 Therefore R = 7 k k= = = 6 8 =.. L 6, f= 6 +, [, ] Let f = { 6 + on[, ]. For n = 6, = /6 =, and { k} 6 k= =,, 5,, 7, 8 }., Therefore L 6 = 5 6k + k= = M 6, f= +, [5, 8] Let f = + on[5, 8]. For n = 6, = 8 5/6 =, and { k }5 k= ={5.5, 5.75, 6.5, 6.75, 7.5, 7.75}. Therefore, M 6 = 5 k + k= = = 7 = 87. April,

7 SECTIO 5. Approimating and Computing Area R 5, f= +, [, ] Let f = + on [, ]. For n = 5, = /5 = 5, and { k} 5 k= {, = 5, 5, 5, 5, }. Therefore R 5 = 5 5 k + k = 5 k= = = L 6, f= +, [, ] Let f = + on [, ]. For n = 6, = /6 =, and { k} 6 k= ={,.5,,.5,,.5, }. Therefore L 6 = 5 k + k = =.5. k= 8. M, f=, [, 5] Let f= on [, 5]. For n =, = 5 / =, and { k } k= ={, 5, 7, 9 }. Therefore, 9. L, f= cos [, π6, π ] Let f= cos on [ π 6, π ]. For n =, M = k k= = π/ π/6 { { k } π k= = 6, π, π, 5π, π }. = = π and Therefore L = π cos k.6. k=. M 5, f= ln, [, ] Let f= ln on [, ]. For n = 5, = /5 = 5, and { k } k= ={6 5, 8 5,, 5, 5 }. Therefore, M 5 = ln 5 k k= = 5 ln ln 8 + ln + ln + ln In Eercises 6, write the sum in summation notation The first term is 7, and the last term is 8 7, so it seems the kth term is k 7. Therefore, the sum is: 8 k 7. k= The first term is +, and the last term is 5 + 5, so it seems that the sum limits are and 5, and the kth term is k + k. Therefore, the sum is: 5 k + k. k= April,

8 57 CHAPTER 5 THE ITEGRAL The first term is +, and the last term is 5 +, so it seems the sum limits are and 5, and the kth term is k +. Therefore, the sum is: n + n 5 k +. k= The first term is + and the last term is n + n, so it seems the summation limits are through n, and the k-th term is k + k. Therefore, the sum is n k= k + k n n + n + The first summand is + +. This shows us n n n + n + = i i + i +. i= 6. e π + e π/ + e π/ + +e π/n The first term is e π/ and the last term is e π/n, so it seems the sum limits are and n and the kth term is e π/k. Therefore, the sum is n e π/k. k= 7. Calculate the sums: 5 5 a 9 b c k i= i= k= a 9 = = 5. Alternativel, 9 = 9 = 95 = 5. i= i= i= b = =. Alternativel, = = 6 =. i= c k = + + = 99. Alternativel, k= k = k k = k= k= k= i= i= = Calculate the sums: a sin j π j= a sin j= 5 b k= c j= jπ = sin π + sin k = + + =. j = + + =. 5 b k k= π = + =. c j j= April,

9 SECTIO 5. Approimating and Computing Area Let b =, b =, b =, and b =. Calculate: a b i b b j b j c kb k i= j= k= a b i = b + b + b = + + =. i= b b j b j = + =. j= c kb k = + + =. k=. Assume that a = 5, a i =, and b i = 7. Calculate: i= i= a a i + b a i i= i= a a i + = a i + = + =. i= i= i= b a i = a i a = 5 = 5. i= i= c a i b i = a i b i = 7 = 9. i=. Calculate j= j= i= i= j. Hint: Write as a difference of two sums and use formula. j = j= j j= j = c a i b i i= + + = 55 = 55.. Calculate j +. Hint: Epand and use formulas. j= j + = j + j + j= j= j= j= = = 9,7. + In Eercises, use linearit and formulas 5 to rewrite and evaluate the sums.. 8j j= 8j = 8 j = = 8, = 5,8. j= j= April,

10 57 CHAPTER 5 THE ITEGRAL. k k= k = k k= k= k= = + = 65 9 = n n=5 5 n=5 n 5 = n 5 n = n= n= =,6,75,95 =,9,5. 6. k= k k= k = k k = k= k= jj j= =,, 5,5,5 = 78,57, jj = j j = j 5 j j= j= j= j= 5 = The power sum formula is usable because jj = jj. 8. 6j + j j= j= j= = 5 5 =,95 =,65. 6j + j = 6 j + j = 6 j j + j j j= j= j= j= j= j= j= = = = ,86 = 6,68. April,

11 SECTIO 5. Approimating and Computing Area m m= m = m= 6 8m + m m m= = 6 8 m + m m m= m= = 6 8 m= m= = 9, +,6 6,5 =, m= 5 + m m= 5 + m = m + 9 m m= m= m= = = = 7.5. In Eercises, use formulas 5 to evaluate the limit. i. lim i= Let s = i= Therefore, lim s =. j. lim j= Let s = j= i. Then, s = i= j. Then s = j= i = i= i = + = +. j = + + = + +. Therefore, lim s =. April,

12 576 CHAPTER 5 THE ITEGRAL. lim i= i i + Let s = i= i i +. Then s = i= = i i + = [ Therefore, lim s =. i. lim i= i Let s = i=. Then s = i= i i= i i + i= i= i= ] + + = +. = + + = + +. Therefore, lim s = = 79. In Eercises 5 5, calculate the limit for the given function and interval. Verif our answer b using geometr. 5. lim R, f= 9, [, ] Let f = 9 on [, ]. Let be a positive integer and set a =, b =, and = b a/ = / = /. Also, let k = a + k = k/, k =,,..., be the right endpoints of the subintervals of [, ]. Then The area under the graph is R = f k = k 9 = 6 k= k= k= lim R = lim = 8. k = 6 + = The region under the graph is a triangle with base and height 8. The area of the region is then 8 = 8, which agrees with the value obtained from the limit of the right-endpoint approimations. 6. lim R, f= + 6, [, ] Let f = + 6on[, ]. Let be a positive integer and set a =, b =, and = b a/ = / = /. Also, let k = a + k = + k/, k =,,..., be the right endpoints of the subintervals of [, ]. Then The area under the graph is R = f k = 9 + 9k k= k= = 7 k= = k= j = lim R = lim + 7 = 8. + April,

13 SECTIO 5. Approimating and Computing Area 577 The region under the graph is a trapezoid with base width and heights 9 and 8. The area of the region is then = 8, which agrees with the value obtained from the limit of the right-endpoint approimations. 7. lim L, f= +, [, ] Let f = + on[, ]. Let >be an integer, and set a =,b =, and = / =. Also, let k = + k = k,k =,,..., be the left endpoints of the subintervals. Then The area under the graph is L = f k = k= = k= + k + = 8 k= =. lim L =. + 8 k= The region under the curve over [, ] is a trapezoid with base width and heights and. From this, we get that the area is + =, which agrees with the answer obtained from the limit of the left-endpoint approimations. 8. lim L, f=, [, ] Let f = on[, ]. Let >be an integer, and set a =,b =, and = / =. Also, let k = a + k = + k,k =,,..., be the left endpoints of the subintervals. Then The area under the graph is L = f k = = 6 = k= k= + 8k + = 6 k= + lim L =. k + The region under the curve over [, ] is a trapezoid with base width and heights and. From this, we get that the area is + =, which agrees with the answer obtained from the limit of the left-endpoint approimations. 9. lim M, f=, [, ] Let f = on [, ]. Let >be an integer and set a =, b =, and = b a/ =. Also, let k = + k = k,k =,,..., be the midpoints of the subintervals of [, ]. Then The area under the curve over [, ] is M = fk = k= k= = k = k= lim M =. k = k k= + =. The region under the curve over [, ] is a triangle with base and height, and thus area, which agrees with the answer obtained from the limit of the midpoint approimations. k= k April,

14 578 CHAPTER 5 THE ITEGRAL 5. lim M, f=, [, 6] Let f= on [, 6]. Let >be an integer and set a =, b = 6, and = b a/ =. Also, let k = a + k = + k,k =,,..., be the midpoints of the subintervals of [, 6]. Then The area under the curve over [, 6] is M = fk = 6k 8 k= k= = 6 k= 6 = 6 6 k= k + k= + + = 6. lim M = 6. The region under the curve over [, 6] consists of a triangle of base and height above the ais and a triangle of base and height below the ais. The area of this region is therefore = 6, which agrees with the answer obtained from the limit of the midpoint approimations. 5. Show, for f= + over [, ], that R = j= j + 6j Then evaluate lim R. Let f= + on [, ]. Let be a positive integer and set a =, b =, and = b a/ = / = /. Also, let j = a + j = j/, j =,,..., be the right endpoints of the subintervals of [, ]. Then R = f j = j + j j= j= = j j= + 8j Continuing, we find Thus, R = = j + 6 j j= = j= lim R = lim = Show, for f= over [, 5], that R = 9j j= + 8j + 8j + Then evaluate lim R. April,

15 SECTIO 5. Approimating and Computing Area 579 Let f= on [, 5]. Let be a positive integer and set a =, b = 5, and = b a/ = 5 / = /. Also, let j = a + j = + j/, j =,,..., be the right endpoints of the subintervals of [, 5]. Then f j = + j + j = + j + 8j + 6j + 8j + 6j and Continuing, we find Thus, = 9j + 8j + 8j +. R = f j = 9j j= j= + 8j + 8j +. R = 768 = 768 = 8 j + 5 j= j= lim R = lim j j + 8 j= j= = 8. In Eercises 5 6, find a formula for R and compute the area under the graph as a limit. 5. f=, [, ] Let f= on the interval [, ]. Then = = and a =. Hence, R = f + j = j j= j= = + + = and 5. f=, [, 5] lim R = lim =. Let f= on the interval [, 5]. Then = 5 = 6 and a =. Hence, R = f + j = 6 + 6j j= j= = 6 j= 7 = 6 7 = j= j j j= April,

16 58 CHAPTER 5 THE ITEGRAL and lim R = lim =. 55. f= 6, [, 5] Let f= 6 on the interval [, 5]. Then = 5 = and a =. Hence, R = f + j = 6 + j = + 7j + 5j j= j= j= = = j= j j j= = and lim R = lim =. 56. f= + 7, [6, ] Let f= + 7 on the interval [6, ]. Then = 6 = 5 and a = 6. Hence, [ R = f6 + j = j j ] j= j= = 5 5j j= + 95j + 78 = 5 = 5 = 5 6 j + 75 j= j= j j= and 5 lim R = lim = f=, [, ] Let f= on the interval [, ]. Then = = and a =. Hence, R = f + j = j j = 8j j= j= j= j = 6 = 6 j j j= j= + + = April,

17 SECTIO 5. Approimating and Computing Area 58 and lim R = lim =. 58. f= +, [, ] Let f= + on the interval [, ]. Then = = and a =. Hence, [ R = f + j = + j + + j ] j= j= = 8j j= 76j + 8j = = and 6 lim = lim = f= +, [a,b] a,b constants with a < b Let f= + on the interval [a,b]. Then = b a. Hence, R = = = fa+ j = j= b a a + + j= b a j= b a b a b a a + + a + j j j= + b a + and = b aa + + b a + b a lim R = lim b aa + + b a + b a 6. f=, [a,b] a,b constants with a<b = b aa + + b a = b + b a + a. Let f= on the interval [a,b]. Then = b a. Hence, b a R = fa+ j = a b a + aj j= j= = a b a + j= ab a = a b a ab a + = a b a + ab a + j + j= + + ab a + b a b a j j= b a + + j b a b a + b a 6 April,

18 58 CHAPTER 5 THE ITEGRAL and lim R = lim a b a + ab a + ab a + b a + b a + b a 6 = a b a + ab a + b a = b a. In Eercises 6 6, describe the area represented b the limits. 6. lim j= The limit j lim R = lim j= j represents the area between the graph of f= and the -ais over the interval [, ]. 6. lim + j j= The limit lim R = lim j= + j represents the area between the graph of f= and the -ais over the interval [, 5] lim e +5j/ j= The limit lim L 5 = lim e +5j/ j= represents the area between the graph of = e and the -ais over the interval [, ]. π π 6. lim sin π + jπ j= The limit π lim j= π sin π + jπ represents the area between the graph of = sin and the -ais over the interval [ π, 5π 6 ]. In Eercises 65 7, epress the area under the graph as a limit using the approimation indicated in summation notation, but do not evaluate. 65. R, f= sin over [,π] Let f= sin over [,π] and set a =, b = π, and = b a / = π/. Then Hence R = f k = π kπ sin. k= k= lim R π = lim sin k= is the area between the graph of f= sin and the -ais over [,π]. kπ April,

19 SECTIO 5. Approimating and Computing Area R, f= over [, 7] Let f= over the interval [, 7]. Then = 7 = 6 and a =. Hence, R = f + j = 6 + j 6 j= j= and lim R 6 = lim j= + j 6 is the area between the graph of f= and the -ais over [, 7]. 67. L, f= + over [7, ] Let f= + over the interval [7, ]. Then = 7 = and a = 7. Hence, L = f7 + j = 7 + j + j= j= and lim L = lim j= 5 + 8j is the area between the graph of f= + and the -ais over [7, ]. 68. L, f= cos over [ π 8, π ] Let f= cos over the interval [ π 8, π ] π. Then = π 8 and L = f j= lim L = lim π 8 + j = π 8 j= π 8 j= is the area between the graph of f= cos and the -ais over [ π 8, π ]. 69. M, f= tan over [, ] = π 8 = π 8 and a = π 8, Hence: π cos 8 + j π 8 π cos 8 + j π 8 Let f= tan over the interval [, ]. Then = = and a =. Hence M = f j= + j = j= tan + j and so lim M = lim j= tan + j is the area between the graph of f= tan and the -ais over [, ]. 7. M, f= over [, 5] Let f= over the interval [, 5]. Then = 5 = and a =. Hence M = f j= + j = j= + j April,

20 58 CHAPTER 5 THE ITEGRAL and so lim M = lim j= + j is the area between the graph of f= and the -ais over [, 5]. j 7. Evaluate lim b interpreting it as the area of part of a familiar geometric figure. j= The limit lim R = lim j= j represents the area between the graph of = f= and the -ais over the interval [, ]. This is the portion of the circular disk + that lies in the first quadrant. Accordingl, its area is π = π. In Eercises 7 7, let f= and let R, L, and M be the approimations for the interval [, ]. 7. Show that R = + +. Interpret the quantit 6 + as the area of a region. 6 The quantit Let f= on [, ]. Let > be an integer and set a =, b = and = =. Then R = f + j = j j= j= = + + = in R = represents the collective area of the parts of the rectangles that lie above the graph of f.it is the error between R and the true area A = Show that L = + 6, M = Then rank the three approimations R, L, and M in order of increasing accurac use Eercise 7. Let f = on [, ]. Let be a positive integer and set a =, b =, and = b a / = /. Let k = a + k = k/, k =,,..., and let k = a + k + = k + /, k =,,...,. Then L = f k = = k= + k= k = k= + 6 k = + 6 April,

21 M = fk = k= = k + k + = k= k= k + k= k= + k= = SECTIO 5. Approimating and Computing Area 585 k + k = The error of R is given b + 6, the error of L is given b + 6 and the error of M is given b. Of the three approimations, R is the least accurate, then L and finall M is the most accurate. 7. For each of R, L, and M, find the smallest integer for which the error is less than.. For R, the error is less than. when: We find an adequate in : + 6 < <. + <.6 <.6, in particular, if > = 5.. Hence R 5 is within. of A. For L, the error is less than. if + 6 <.. We solve this equation for : 6 <. 6 <. <.6 <.6 +, which is satisfied if > = Therefore, L 5 is within. units of A. For M, the error is given b, so the error is less than. if <. < 9. < Therefore, M is within. units of the correct answer. In Eercises 75 8, use the Graphical Insight on page 9 to obtain bounds on the area. 75. Let A be the area under f= over [, ]. Prove that.5 A.77 b computing R and L. Eplain our reasoning. For n =, = = and { i} i= ={ + i }={,,,, }. Therefore, R = f i = i= April,

22 586 CHAPTER 5 THE ITEGRAL L = f i = i= In the plot below, ou can see the rectangles whose area is represented b L under the graph and the top of those whose area is represented b R above the graph. The area A under the curve is somewhere between L and R,so.58 A.768. L, R and the graph of f. 76. Use R 5 and L 5 to show that the area A under = over [, ] satisfies.8 A.. Let f= over the interval [, ]. Because f is a decreasing function over this interval, it follows that R A L for all. Taking = 5, we have = /5 and R 5 = =.885. Moreover, L 5 = =.. Thus,.8 <R 5 A L 5 < Use R and L to show that the area A under the graph of = sin over [, π ] satisfies.79 A.9. Let f = sin. fis increasing over the interval [,π/], so the Insight on page 9 applies, which indicates that L A R. For n =, = π/ = π 8 and { i} i= ={ + i } i= ={, π 8, π, π 8, π }. From this, L = π f i.79, R = π f i i= i= Hence A is between.79 and.9. Left and Right endpoint approimations to A. 78. Show that the area A under f= over [, 8] satisfies A Let f =, 8. Since f is decreasing, the left endpoint approimation L 7 overestimates the true area between the graph of f and the -ais, whereas the right endpoint approimation R 7 underestimates it. Accordingl, = R 7 <A<L 7 = Left endpoint approimation, n = Right endpoint approimation, n = April,

23 SECTIO 5. Approimating and Computing Area Show that the area A under = / over [, ] satisfies L A R for all. Use a computer algebra sstem to calculate L and R for = and, and determine A to two decimal places. On [, ], f= / is an increasing function; therefore, L A R for all. We find while L = and R =.899, L =.7977 and R =.875. Thus, A =.8 to two decimal places. 8. Show that the area A under = / + over [, ] satisfies R A L for all. Determine A to at least three decimal places using a computer algebra sstem. Can ou guess the eact value of A? On [, ], the function f = / + is decreasing, so R A L for all. From the values in the table below, we find A =. to three decimal places. It appears that the eact value of A is π. R L In this eercise, we evaluate the area A under the graph of = e over [, ] [Figure 9A] using the formula for a geometric sum valid for r : + r + r + +r = r j = r r j= 8 a Show that L = e j/. j= b Appl Eq. 8 with r = e / e to prove L = e /. c Compute A = lim L using L Hôpital s Rule. e = e = ln A B e A FIGURE 9 B a Let f= e on [, ]. With n =, = / = / and for j =,,,...,. Therefore, b Appling Eq. 8 with r = e /, we have j = a + j = j L = f j = j= j= e j/. L = e / e e / = e /. April,

24 588 CHAPTER 5 THE ITEGRAL Therefore, A = lim L = e lim e /. c Using L Hôpital s Rule, A = e lim e / = e lim = e lim e/ e / = e. 8. Use the result of Eercise 8 to show that the area B under the graph of f = ln over [,e] is equal to. Hint: Relate B in Figure 9B to the area A computed in Eercise 8. Because = ln and = e are inverse functions, we note that if the area B is reflected across the line = and then combined with the area A, we create a rectangle of width and height e. The area of this rectangle is therefore e, and it follows that the area B is equal to e minus the area A. Using the result of Eercise 8, the area B is equal to e e =. Further Insights and Challenges 8. Although the accurac of R generall improves as increases, this need not be true for small values of. Draw the graph of a positive continuous function fon an interval such that R is closer than R to the eact area under the graph. Can such a function be monotonic? Let δ be a small positive number less than. In the figures below, δ =. But imagine δ being ver tin. Define fon [, ] b if < δ δ f= δ if δ < δ δ if < + δ if + δ Then f is continuous on [, ]. Again, just look at the figures. The eact area between f and the -ais is A = bh = δ = δ. For δ =, we have A = 9. With R =, the absolute error is E = R A = δ = δ. For δ =, this absolute error is E =. With R =, the absolute error is E = R A = δ = δ = δ. For δ =, we have E = 5. Accordingl, R is closer to the eact area A than is R. Indeed, the tinier δ is, the more dramatic the effect. For a monotonic function, this phenomenon cannot occur. Successive approimations from either side get progressivel more accurate Graph of f Right endpt appro, n = Right endpt appro, n = Draw the graph of a positive continuous function on an interval such that R and L are both smaller than the eact area under the graph. Can such a function be monotonic? In the plot below, the area under the saw-tooth function fis, whereas L = R =. Thus L and R are both smaller than the eact area. Such a function cannot be monotonic; if fis increasing, then L underestimates and R overestimates the area for all, and, if fis decreasing, then L overestimates and R underestimates the area for all. April,

25 SECTIO 5. Approimating and Computing Area 589 Left/right-endpoint approimation, n = 85. Eplain graphicall: The endpoint approimations are less accurate when f is large. When f is large, the graph of f is steeper and hence there is more gap between f and L or R. Recall that the top line segments of the rectangles involved in an endpoint approimation constitute a piecewise constant function. If f is large, then f is increasing more rapidl and hence is less like a constant function. Smaller f' Larger f' 86. Prove that for an function fon [a,b], R L = b a f b fa 9 For an f continuous or not on I = [a,b], partition I into equal subintervals. Let = b a/ and set k = a + k, k =,,... Then we have the following approimations to the area between the graph of f and the -ais: the left endpoint approimation L = k= f k and right endpoint approimation R = k= f k. Accordingl, R L = f k f k In other words, R L = b a k= k= = f + f k f f k k= = f f = b a fb fa. k= fb fa. 87. In this eercise, we prove that lim R and lim L eist and are equal if fis increasing [the case of fdecreasing is similar]. We use the concept of a least upper bound discussed in Appendi B. a Eplain with a graph wh L R M for all,m. b B a, the sequence {L } is bounded, so it has a least upper bound L. B definition, L is the smallest number such that L L for all. Show that L R M for all M. c According to b, L L R for all. Use Eq. 9 to show that lim L = L and lim R = L. a Let fbe positive and increasing, and let and M be positive integers. From the figure below at the left, we see that L underestimates the area under the graph of = f, while from the figure below at the right, we see that R M overestimates the area under the graph. Thus, for all,m, L R M. April,

26 59 CHAPTER 5 THE ITEGRAL b Because the sequence {L } is bounded above b R M for an M, each R M is an upper bound for the sequence. Furthermore, the sequence {L } must have a least upper bound, call it L. B definition, the least upper bound must be no greater than an other upper bound; consequentl, L R M for all M. c Since L L R, R L R L,so R L R L. From this, lim R L lim R L. B Eq. 9, lim R L = lim b af b fa =, so lim R L R L =, hence lim R = L. Similarl, L L =L L R L,so This gives us that L L R L = b a f b f a. lim L L lim b af b fa =, so lim L = L. This proves lim L = lim R = L. 88. Use Eq. 9 to show that if fis positive and monotonic, then the area A under its graph over [a,b] satisfies R A b a fb fa Let fbe continuous, positive, and monotonic on [a,b]. Let A be the area between the graph of f and the -ais over [a,b]. For specificit, sa f is increasing. The case for f decreasing on [a,b] is similar. As noted in the tet, we have L A R. B Eercise 86 and the fact that A lies between L and R, we therefore have R A R L = b a fb fa. Hence R A b a b a fb fa = fb fa, where fb fa= fb fa because f is increasing on [a,b]. In Eercises 89 and 9, use Eq. to find a value of such that R A < for the given function and interval. 89. f=, [, ] Let f= on [, ]. Then b =, a =, and R A f f = =. We need <, which gives >,. Thus R, A < for f= on [, ]. 9. f= 9, [, ] Let f= 9 on [, ]. Then b =,a =, and R A b a fb fa = = 9. We need 9 <, which gives >9,. Thus R 9, A < for f= 9 on [, ]. 9. Prove that if fis positive and monotonic, then M lies between R and L and is closer to the actual area under the graph than both R and L. Hint: In the case that fis increasing, Figure shows that the part of the error in R due to the ith rectangle is the sum of the areas A + B + D, and for M it is B E. On the other hand, A E. April,

27 SECTIO 5. Approimating and Computing Area 59 A B C D E F i midpoint i FIGURE Suppose fis monotonic increasing on the interval [a,b], = b a, { k } k= = {a,a +, a +,...,a+, b} and { { } a + a + k k= =, a + + a +,..., } a + + b. ote that i <i < i+ implies f i <fi <f i+ for all i<because fis monotone increasing. Then L = b a f k < M = b a f k < R = b a f k k= k= k= Similarl, if fis monotone decreasing, L = b a f k > M = b a k= k= fk > R = b a f k Thus, if fis monotonic, then M alwas lies in between R and L. ow, as in Figure, consider the tpical subinterval [ i, i ] and its midpoint i. We let A, B, C, D, E, and F be the areas as shown in Figure. ote that, b the fact that i is the midpoint of the interval, A = D + E and F = B + C. Let E R represent the right endpoint approimation error = A + B + D, let E L represent the left endpoint approimation error = C + F + E and let E M represent the midpoint approimation error = B E. If B>E, then E M = B E. In this case, so E R >E M, while E R E M = A + B + D B E = A + D + E>, E L E M = C + F + E B E = C + B + C + E B E = C + E >, so E L >E M. Therefore, the midpoint approimation is more accurate than either the left or the right endpoint approimation. If B<E, then E M = E B. In this case, so that E R >E M while E R E M = A + B + D E B = D + E + D E B = D + B>, E L E M = C + F + E E B = C + F + B>, so E L >E M. Therefore, the midpoint approimation is more accurate than either the right or the left endpoint approimation. If B = E, the midpoint approimation is eactl equal to the area. Hence, for B<E, B>E,orB = E, the midpoint approimation is more accurate than either the left endpoint or the right endpoint approimation. k= April,

28 59 CHAPTER 5 THE ITEGRAL 5. The Definite Integral Preliminar Questions 5. What is d [the function is f= ]? 5 5 d = d = 5 =. 7. Let I = fd, where fis continuous. State whether true or false: a I is the area between the graph and the -ais over [, 7]. b If f, then I is the area between the graph and the -ais over [, 7]. c If f, then I is the area between the graph of fand the -ais over [, 7]. a False. b a fd is the signed area between the graph and the -ais. b True. c True. π. Eplain graphicall: cos d=. Because cosπ = cos, the negative area between the graph of = cos and the -ais over [ π,π] eactl cancels the positive area between the graph and the -ais over [, π ]. 5. Which is negative, 8 d or 8 d? 5 5 Because 5 =, 8 d is negative. Eercises In Eercises, draw a graph of the signed area represented b the integral and compute it using geometr.. d The region bounded b the graph of = and the -ais over the interval [, ] consists of two right triangles. One has area 6 = 9 below the ais, and the other has area 6 = 9 above the ais. Hence, d= 9 9 = d The region bounded b the graph of = + and the -ais over the interval [, ] consists of a single right triangle of area 5 = 5 above the ais. Hence, + d = April,

29 SECTIO 5. The Definite Integral d The region bounded b the graph of = + and the -ais over the interval [, ] consists of two right triangles. One has area = below the ais, and the other has area 7 7 = 9 6 above the ais. Hence, + d = 9 6 = d The region bounded b the graph of = and the -ais over the interval [, ] is a rectangle of area = above the ais. Hence, d = d 6 The region bounded b the graph of = 7 and the -ais over the interval [6, 8] consists of two right triangles. One triangle has area = above the ais, and the other has area = below the ais. Hence, 8 7 d = 6 = π/ 6. sin d π/ The region bounded b the graph of = sin and the -ais over the interval [ π, π ] consists of two parts of equal area, one above the ais and the other below the ais. Hence, π/ sin d=. π/.5.5 April,

30 59 CHAPTER 5 THE ITEGRAL d The region bounded b the graph of = 5 and the -ais over the interval [, 5] is one-quarter of a circle of radius 5. Hence, 5 5 d = π5 = 5π d The region bounded b the graph of = and the -ais over the interval [, ] consists of two right triangles, both above the ais. One triangle has area =, and the other has area = 9. Hence, d = 9 + =. 9. d The region bounded b the graph of = and the -ais over the interval [, ] is a triangle above the ais with base and height. Consequentl, d = = d The region bounded b the graph of = + and the -ais over the interval [, 5] consists of a triangle below the ais with base and height, a triangle above the ais of base and height and a triangle below the ais of base and height. Consequentl, 5 + d = + = 5. April,

31 SECTIO 5. The Definite Integral 595. Calculate 8 d in two was: a As the limit lim R b B sketching the relevant signed area and using geometr Let f= 8 over [, ]. Consider the integral fd = 8 d. a Let be a positive integer and set a =, b =, = b a / = /. Also, let k = a + k = k/, k =,,..., be the right endpoints of the subintervals of [, ]. Then R = = k= f k = 8 8 k = 8 k k= k= k= = 5. + Hence lim R = lim 5 =. b The region bounded b the graph of = 8 and the -ais over the interval [, ] consists of two right triangles. One triangle has area 88 = above the ais, and the other has area = below the ais. Hence, 8 d = = Calculate 8d in two was: As the limit lim R and using geometr. Let f= 8 over [, ]. Consider the integral fd = 8d. Let be a positive integer and set a =, b =, = b a / = 5/. Then k = a + k = + 5k/, k =,,..., are the right endpoints of the subintervals of [, ]. Then R = f k = 5 k= k= = k 8 = 6 + k= k k= = = +. Hence lim R = lim + 5 =. The region bounded b the graph of = 8 and the -ais over the interval [, ] consists of a triangle below the ais with base and height and a triangle above the ais with base and height 8. Hence, 8d = + 8 =. 5 5 April,

32 596 CHAPTER 5 THE ITEGRAL In Eercises and, refer to Figure. = f 6 FIGURE The two parts of the graph are semicircles. 6. Evaluate: a fd b fd Let fbe given b Figure. a The definite integral fdis the signed area of a semicircle of radius which lies below the -ais. Therefore, fd = π = π. b The definite integral 6 fd is the signed area of a semicircle of radius which lies below the -ais and a semicircle of radius which lies above the -ais. Therefore, 6 fd = π π = π. 6. Evaluate: a fd b f d Let fbe given b Figure. a The definite integral fd is the signed area of one-quarter of a circle of radius which lies below the -ais and one-quarter of a circle of radius which lies above the -ais. Therefore, fd = π π = π. b The definite integral 6 f d is the signed area of one-quarter of a circle of radius and a semicircle of radius, both of which lie above the -ais. Therefore, 6 f d = π + π = 9π. In Eercises 5 and 6, refer to Figure 5. = gt 5 FIGURE 5 t 5 5. Evaluate gt dt and gt dt. The region bounded b the curve = g and the -ais over the interval [, ] is comprised of two right triangles, one with area below the ais, and one with area above the ais. The definite integral is therefore equal to =. The region bounded b the curve = g and the -ais over the interval [, 5] is comprised of another two right triangles, one with area above the ais and one with area below the ais. The definite integral is therefore equal to. April,

33 SECTIO 5. The Definite Integral 597 a c 6. Find a, b, and c such that gt dt and gt dt are as large as possible. b a To make the value of gt dt as large as possible, we want to include as much positive area as possible. c This happens when we take a =. ow, to make the value of gt dt as large as possible, we want to make sure to b include all of the positive area and onl the positive area. This happens when we take b = and c =. 7. Describe the partition P and the set of sample points C for the Riemann sum shown in Figure 6. Compute the value of the Riemann sum FIGURE 6 The partition P is defined b = < = < =.5 < =. < = 5 The set of sample points is given b C ={c =.5,c =,c =,c =.5}. Finall, the value of the Riemann sum is = Compute Rf, P, C for f= + for the partition P and the set of sample points C in Figure 6. Rf, P, C = f.5 + f.5 + f..5 + f.55. = = In Eercises 9, calculate the Riemann sum Rf, P, C for the given function, partition, and choice of sample points. Also, sketch the graph of f and the rectangles corresponding to Rf, P, C. 9. f=, P ={,.,.5, }, C ={.,.,.9} Let f=. With we get P ={ =, =., =.5, = } and C ={c =.,c =.,c =.9}, Rf, P, C = fc + fc + fc Here is a sketch of the graph of f and the rectangles. = = f= +, P ={,,,, 8}, C ={,,, 5} Let f= +. With P ={ =, =, =, =, = 8} and C ={c =,c =,c =,c = 5}, we get Rf, P, C = fc + fc + fc + fc = = 7. April,

34 598 CHAPTER 5 THE ITEGRAL Here is a sketch of the graph of f and the rectangles f= +, P ={,,.5, 5}, C ={,.5, 5} Let f= +. With P ={ =, =, =.5, = 5} and C ={c =,c =.5,c = 5}, we get Rf, P, C = fc + fc + fc = =.65. Here is a sketch of the graph of f and the rectangles f= sin, P = {, π 6, π, π }, C ={.,.7,.} Let f= sin. With { P = =, = π 6, = π, = π } and C ={c =.,c =.7,c =.}, we get Rf, P, C = fc + fc + fc π = 6 sin. + Here is a sketch of the graph of f and the rectangles. π π 6 sin.7 + π π sin. = In Eercises 8, sketch the signed area represented b the integral. Indicate the regions of positive and negative area. 5. d Here is a sketch of the signed area represented b the integral 5 d. 5 April,

35 π/. tan d π/ Here is a sketch of the signed area represented b the integral π/ π/ tan d. SECTIO 5. The Definite Integral π 5. sin d π Here is a sketch of the signed area represented b the integral π π. sin d π 6. sin d Here is a sketch of the signed area represented b the integral π sin d ln d / Here is a sketch of the signed area represented b the integral / ln d tan d Here is a sketch of the signed area represented b the integral tan d April,

36 6 CHAPTER 5 THE ITEGRAL In Eercises 9, determine the sign of the integral without calculating it. Draw a graph if necessar. 9. d The integrand is alwas positive. The integral must therefore be positive, since the signed area has onl positive part.. d B smmetr, the positive area from the interval [, ] is cancelled b the negative area from [, ]. With the interval [, ] contributing more negative area, the definite integral must be negative. π. sin d As ou can see from the graph below, the area below the ais is greater than the area above the ais. Thus, the definite integral is negative π sin d From the plot below, ou can see that the area above the ais is bigger than the area below the ais, hence the integral is positive In Eercises, use properties of the integral and the formulas in the summar to calculate the integrals.. 6t dt 6t dt = 6 tdt dt = 6 = d + 7d = d+ 7 d = d+ d + 7 = d d + 5 = + 5 = d 9 B formula 5, d = 9 =. April,

37 SECTIO 5. The Definite Integral d 5 5 d = d d = 5 = u u du u u du = u du udu= = =. / d / / / + 6d = d + 6 d = t + t + dt First, write = + = 5. Then,. 9 d First write 7t + t + dt = 7t + t + dt + 7t + t + dt = 7t + t + dt + 7t + t + dt 7t + t + dt = = = Then, 9 d = 9 d+ 9 d = 9 d+ 9 d. 9 d = 9 8 = = d a First, b + d = b d + b d= b + b. Therefore a + d = + d + + d = + d + d a a = + a + a = a a April,

38 6 CHAPTER 5 THE ITEGRAL a. d a a a a d = d d = a a a = a6 a. In Eercises 7, calculate the integral, assuming that 5 5 fd = 5, g d = 5. f + g d f + g d = fd+ g d = 5 + = f g d 5 f 5 g d = fd 5 g d = 5 = g d 5 5 g d = g d = f d f d = fd d= 5 5 = Is it possible to calculate gfd from the information given? It is not possible to calculate 5 gfd from the information given. 8. Prove b computing the limit of right-endpoint approimations: b d = b 9 Let f=, a = and = b a/ = b/. Then R = f k = b k b k= k= = b k = b k= b Hence d = lim R b = lim + b + b = b. In Eercises 9 5, evaluate the integral using the formulas in the summar and Eq d B Eq. 9, d = = d d = d d = =. + + = b + b + b. April,

39 SECTIO 5. The Definite Integral 6 5. d d = d d = = d Appling the linearit of the definite integral, Eq. 9, the formula from Eample and the formula for the definite integral of a constant: + d = d d+ d = + = d + 8d = d + 8 d = + 8 = + 8 = 7 5. d d = d+ d = d d = d d d + d = + = = 6. In Eercises 55 58, calculate the integral, assuming that fd =, fd =, 55. fd fd = fd+ fd = + 7 = fd fd = fd fd = =. 57. fd fd = fd = fd From Eercise 55, fd = 8. Accordingl, fd = 7 fd = fd fd = 8 =. April,

40 6 CHAPTER 5 THE ITEGRAL In Eercises 59 6, epress each integral as a single integral fd+ fd 7 7 fd+ fd = fd fd fd fd fd = fd+ fd fd = fd fd fd fd fd = fd+ fd fd = fd fd+ fd fd+ fd = fd+ fd+ fd = fd b In Eercises 6 66, calculate the integral, assuming that f is integrable and fd = b for all b> fd 5 fd = 5 = fd 5 5 fd = fd fd = = f d f d = fd d = 6 6 = fd / / fd = fd = =. / b b 67. Eplain the difference in graphical interpretation between fd and f d. a a When f takes on both positive and negative values on [a,b], b a fd represents the signed area between f and the -ais, whereas b a f d represents the total unsigned area between f and the -ais. An negativel signed areas that were part of b a fdare regarded as positive areas in b a f d. Here is a graphical eample of this phenomenon. Graph of f Graph of f April,

41 SECTIO 5. The Definite Integral Use the graphical interpretation of the definite integral to eplain the inequalit b b fd a f d a where fis continuous. Eplain also wh equalit holds if and onl if either f for all or f for all. Let A + denote the unsigned area under the graph of = f over the interval [a,b] where f. Similarl, let A denote the unsigned area when f<. Then b fd = A + A. a Moreover, b b fd a A + + A = f d. a Equalit holds if and onl if one of the unsigned areas is equal to zero; in other words, if and onl if either f for all or f for all. 69. Let f=. Find an interval [a,b] such that b fd a = and b f d = a If a>, then f for all [a,b], so b b fd a = f d a b the previous eercise. We find a similar result if b<. Thus, we must have a< and b>. ow, Because then b f d = a a + b. b fd = a b a, b fd a = b a. If b >a, then a + b = and ield a = and b =. On the other hand, if b <a, then a + b = and b a = a b = ield a = and b =. π π 7. Evaluate I = sin dand J = cos das follows. First show with a graph that I = J. Then prove that I + J = π. The graphs of f = sin and g = cos are shown below at the left and right, respectivel. It is clear that the shaded areas in the two graphs are equal, thus π π I = sin d= cos d= J. ow, using the fundamental trigonometric identit, we find π π I + J = sin + cos d = d = π. Combining this last result with I = J ields I = J = π. April,

42 66 CHAPTER 5 THE ITEGRAL In Eercises 7 7, calculate the integral d Over the interval, the region between the curve and the interval [, 6] consists of two triangles above the ais, each of which has height and width, and so area 9. The total area, hence the definite integral, is Alternatel, 6 6 d = d + d 6 6 = d d+ d d d = = d The area between and the ais consists of two triangles above the -ais, each with width and height, and hence with area. The total area, and hence the definite integral, is Alternatel, d = d + d = d d d + d d d = + =. 7. d { = <. April,

43 SECTIO 5. The Definite Integral 67 Therefore, d = d + d = d + d = + =. 7. d = { <. Therefore, d = d+ d = d d + d d d = + 8 =. 75. Use the Comparison Theorem to show that 5 d d, d 5 d On the interval [, ], 5, so, b Theorem 5, 5 d d. On the other hand, 5 for [, ], so, b the same Theorem, d 5 d. 76. Prove that 6 d. On the interval [, 6], 6, so, b Theorem 5, 6 = 6 6 d d. On the other hand, on the interval [, 6], so 6 6 d d = 6 =. Therefore 6 d, as desired. 77. Prove that.98.. sin d.96. Hint: Show that.98 sin.96 for in [.,.]. For π 6.5, we have d d sin = cos >. Hence sin is increasing on [.,.]. Accordingl, for.., we have m = sin. sin sin = M Therefore, b the Comparison Theorem, we have = m.. = md sin d Md= M.. = April,

44 68 CHAPTER 5 THE ITEGRAL π/ 78. Prove that.77 cos d.6. π/8 cos is decreasing on the interval [π/8,π/]. Hence, for π/8 π/, cosπ/ cos cosπ/8. Since cosπ/ = /, Since cosπ/8.9,.77 π 8 π/ = π/8 d π/ cos d. π/8 π/ π/ cos d.9 d = π.9.6. π/8 π/8 8 Therefore.77 π/ π/8 cos.6. π/ sin 79. Prove that d π/. Let f= sin. As we can see in the sketch below, fis decreasing on the interval [π/,π/]. Therefore f fπ/ for all in [π/,π/]. fπ/ = π, so: π/ π/ sin π/ d π/ π d = π π =. /p /p = sin d 8. Find upper and lower bounds for 5 +. Let p/ p/ f= 5 +. fis decreasing for on the interval [, ], sof f f for all in [, ]. f = and f =,so d fd d fd. 8. Suppose that f g on [a,b]. B the Comparison Theorem, b a fd b a g d. Is it also true that f g for [a,b]? If not, give a countereample. The assertion f g is false. Consider a =, b =, f=, g =. f g for all in the interval [, ], but f = while g = for all. 8. State whether true or false. If false, sketch the graph of a countereample. b a If f>, then fd >. a b b If fd >, then f>. a April,

45 SECTIO 5. The Definite Integral 69 a It is true that if f> for [a,b], then b a fd >. b It is false that if b a fd >, then f> for [a,b]. Indeed, in Eercise, we saw that + d = 7.5 >, et f = <. Here is the graph from that eercise. 6 Further Insights and Challenges 8. Eplain graphicall: If fis an odd function, then a fd =. a If f is an odd function, then f = ffor all. Accordingl, for ever positivel signed area in the right half-plane where f is above the -ais, there is a corresponding negativel signed area in the left half-plane where f is below the -ais. Similarl, for ever negativel signed area in the right half-plane where f is below the -ais, there is a corresponding positivel signed area in the left half-plane where f is above the -ais. We conclude that the net area between the graph of f and the -ais over [ a,a] is, since the positivel signed areas and negativel signed areas cancel each other out eactl. 8. Compute sinsinsin + d. Let f= sinsinsin +. sin is an odd function, while sin is an even function, so: f = sinsin sin + = sin sinsin + = sinsinsin + = f. Therefore, fis an odd function. The function is odd and the interval is smmetric around the origin so, b the previous eercise, the integral must be zero. 85. Let k and b be positive. Show, b comparing the right-endpoint approimations, that b k d = b k+ k d Let k and b be an positive numbers. Let f = k on [,b]. Since f is continuous, both b fd and fd eist. Let be a positive integer and set = b / = b/. Let j = a + j = bj/, j =,,..., be the right endpoints of the subintervals of [,b]. Then the right-endpoint approimation to b fd = b k d is R = f j = b bj k = b k+ j= j= k+ j k. In particular, if b = above, then the right-endpoint approimation to fd = k d is S = f j = j k = j= j= k+ j= j= j k = b k+ R April,

46 6 CHAPTER 5 THE ITEGRAL In other words, R = b k+ S. Therefore, b k d = lim R = lim bk+ S = b k+ lim S = b k+ k d. 86. Verif for b b interpreting in terms of area: b d = b b + sin b The function f= is the quarter circle of radius in the first quadrant. For b, the area represented b the integral b d can be divided into two parts. The area of the triangular part is b b using the Pthagorean Theorem. The area of the sector with angle θ where sin θ = b, is given b θ. Thus b d = b b + θ = b b + sin b. θ b 87. Suppose that f and g are continuous functions such that, for all a, a a fd = g d a a Give an intuitive argument showing that f = g. Eplain our idea with a graph. Let c = b. Since b<, c>, so b Eq. 5, c d = c. Furthermore, is an even function, so smmetr of the areas gives c d = d. c Finall, b c c d = d = d = d = c c = b. 88. Theorem remains true without the assumption a b c. Verif this for the cases b<a<cand c<a<b. The additivit propert of definite integrals states for a b c, we have c a fd = b a fd + cb fd. Suppose that we have b<a<c. B the additivit propert, we have c b fd = a b fd + c a fd. Therefore, c a fd = c b fd a b fd = b a fd+ c b fd. ow suppose that we have c<a<b. B the additivit propert, we have b c fd = a c fd+ b a fd. Therefore, c a fd = a c fd = b a fd b c fd = b a fd+ c b fd. Hence the additivit propert holds for all real numbers a, b, and c, regardless of their relationship amongst each other. April,

47 SECTIO 5. The Fundamental Theorem of Calculus, Part I 6 5. The Fundamental Theorem of Calculus, Part I Preliminar Questions. Suppose that F = fand F =, F = 7. a What is the area under = fover [, ] if f? b What is the graphical interpretation of F F if ftakes on both positive and negative values? a If f over [, ], then the area under = fis F F = 7 =. b If ftakes on both positive and negative values, then F F gives the signed area between = fand the -ais.. Suppose that fis a negative function with antiderivative F such that F = 7 and F =. What is the area a positive number between the -ais and the graph of fover [, ]? fd represents the signed area bounded b the curve and the interval [, ]. Since fis negative on [, ], fd is the negative of the area. Therefore, if A is the area between the -ais and the graph of f, we have: A = fd = F F = 7 = =.. Are the following statements true or false? Eplain. a FTC I is valid onl for positive functions. b To use FTC I, ou have to choose the right antiderivative. c If ou cannot find an antiderivative of f, then the definite integral does not eist. a False. The FTC I is valid for continuous functions. b False. The FTC I works for an antiderivative of the integrand. c False. If ou cannot find an antiderivative of the integrand, ou cannot use the FTC I to evaluate the definite integral, but the definite integral ma still eist. 9. Evaluate f d where fis differentiable and f = f9 =. 9 Because f is differentiable, f d = f9 f = =. Eercises In Eercises, sketch the region under the graph of the function and find its area using FTC I.. f=, [, ] We have the area A = d = =. April,

48 6 CHAPTER 5 THE ITEGRAL. f=, [, ] Let A be the area indicated. Then: A = d = d d =. f=, [, ] = = We have the area [. f= cos,, π ] A = d = = + = Let A be the shaded area. Then π/ π/ A = cos d= sin = =. In Eercises 5, evaluate the integral using FTC I d 6 d= 6 = 6 = d 9 9 d = = 9 = d 9 d = = =. April,

49 SECTIO 5. The Fundamental Theorem of Calculus, Part I 6 8. u du u du = u = = d 5 + d = 6 + = = d d = + 6 = =.. t 6t dt t 6t dt = t t 8 = 5 = 7.. 5u + u u du 5u + u u du = u 5 + u u = + = 8.. d d= / d = / = / / = / d 8 / d = 8 7 7/ = 8 8 = t / dt /6 t / dt = /6 5 t5/ = /6 5 =. 6. t 5/ dt t 5/ dt = 7 t7/ = 8 = dt 7. t dt t = t dt = t = + =. 8. d d = = + = / d 8 / d = 8 d = = + 6 =. / / April,

50 6 CHAPTER 5 THE ITEGRAL. d. d 9. t / dt 9 7 t +. dt t d = = + = 8. d = + 8 = + + = 6. t / dt = t / 9 = 9 / / =. 7 t + t dt = 7 t / + t / dt = = t/ + t / 7 + = 6 8. t / 8t /. 8/7 t dt t / 8t / 8/7 t dt = t / 8t 5/ dt 8/7 = t / + t / = = 5. 8/7 π/ 5. sin θdθ π/ π/ π/ sin θdθ= cos θ = π/ π/ + =. π 6. sin d π π π sin d= cos = =. π π π/ 7. cos θ dθ π/ π/ cos θ dθ = sin θ =. 5π/8 8. cos d π/ 5π/8 cos d= 5π/8 π/ sin = π/ sin 5π sin π =. π/6 9. sec t π dt 6 π/6 sec t π dt = 6 tan t π π/6 = + = 6. April,

51 SECTIO 5. The Fundamental Theorem of Calculus, Part I 65 π/6. sec θ tan θdθ π/6 π/6 sec θ tan θdθ= sec θ = sec π 6 sec =. π/. csc 5 cot 5d π/ π/ csc 5 cot 5d= π/ π/ 5 csc 5 = = π/ 5 5. π/. csc 7d π/8 π/ csc 7d= π/ π/8 7 cot 7 = π/8 7 cot π + 7 cot π = 7.. e d e d = e = e. 5. e d 5 e d = 5 e = e + e. 5. e 6t dt e 6t dt = 6 e 6t = 6 e e = 6 e e e t dt e t dt = et = e9 e5. d 7. d = ln = ln ln = ln 5. d 8. d = ln = ln ln =ln = ln. dt 9. t + dt = ln t + t + = ln ln = ln. dt. 5t + dt 5t + = ln 5t + 5 = 5 ln 5 ln 9 = ln e d 9e d = e = 6 e 6 = e 6 9. April,

52 66 CHAPTER 5 THE ITEGRAL 6. + d d = + ln = 8 + ln 6 + ln = 6 + ln. In Eercises 8, write the integral as a sum of integrals without absolute values and evaluate.. d d = d + d= + = + = d 5 5 d = d + d = = = d d = d+ d = + = + + = d d = d+ d = = + 9 =. + π 7. cos d π π/ π π/ π cos d = cos d+ cos d = sin sin = =. π/ π/ d d = d 5 = + d+ + d+ + d April,

53 = + = + = 8. SECTIO 5. The Fundamental Theorem of Calculus, Part I In Eercises 9 5, evaluate the integral in terms of the constants. b 9. d b d = b = b = b for an number b. a 5. d b a d = a b 5 5 = b 5 a5 5 b5 for an numbers a, b. b 5. 5 d b 5 d = b 6 6 = 6 b6 6 6 = 6 b6 for an number b. 5. t + tdt t + tdt = t + t = + + =. 5a 5. a b 5. b d d 5a a d 5a = ln = ln 5a ln a =ln 5. a b d b = ln b b 55. Calculate fd, where = ln b ln b =ln b. f= { for for > fd = fd+ fd = d+ d = + = + = = 77. April,

54 68 CHAPTER 5 THE ITEGRAL π 56. Calculate fd, where f= { cos cos sin for π for >π π π π π π fd = fd+ fd = cos d+ cos sin d π π π = sin + sin + π cos π = =. 57. Use FTC I to show that n d = ifn is an odd whole number. Eplain graphicall. We have n d = n+ n + = n+ n + n+ n +. Because n is odd, n + is even, which means that n+ = n+ =. Hence n+ n + n+ = n + n + n + =. Graphicall speaking, for an odd function such as shown here, the positivel signed area from = to = cancels the negativel signed area from = to = Plot the function f= sin. Find the positive root of fto three places and use it to find the area under the graph of fin the first quadrant. The graph of f= sin is shown below at the left. In the figure below at the right, we zoom in on the positive root of fand find that, to three decimal places, this root is approimatel =.76. The area under the graph of fin the first quadrant is then.76 sin d = cos.76 = cos Calculate F given that F = and F =. Hint: Epress F F as a definite integral. B FTC I, F F = d = Therefore F = F + = + =. = April,

55 SECTIO 5. The Fundamental Theorem of Calculus, Part I Calculate G6, where dg/dt = t / and G9 = 5. B FTC I, 6 G6 G9 = t / dt = 6 / 9 / = 9 Therefore G6 = 5 + =. 6. Does n d get larger or smaller as n increases? Eplain graphicall. Let n and consider n d. ote: for n<the integrand n as +, so we eclude this possibilit. ow n d = n + n+ = n + n+ n + n+ = n +, which decreases as n increases. Recall that n d represents the area between the positive curve f = n and the -ais over the interval [, ]. Accordingl, this area gets smaller as n gets larger. This is readil evident in the following graph, which shows curves for several values of n. / / 8 6. Show that the area of the shaded parabolic arch in Figure 8 is equal to four-thirds the area of the triangle shown. a a + b FIGURE 8 Graph of = ab. b We first calculate the area of the parabolic arch: b b b ab d = a b d = a b + abd a a a = a b b + ab a = a b b + 6ab 6 a = b ab b + 6ab a a ba + 6a b 6 = b + ab a + a b 6 = a + ab a b b = 6 6 b a. The indicated triangle has a base of length b a and a height of a + b a b a + b b a =. Thus, the area of the triangle is b a b a = 8 b a. April,

56 6 CHAPTER 5 THE ITEGRAL Finall, we note that as required. 6 b a = 8 b a, Further Insights and Challenges 6. Prove a famous result of Archimedes generalizing Eercise 6: For r<s, the area of the shaded region in Figure 9 is equal to four-thirds the area of triangle ACE, where C is the point on the parabola at which the tangent line is parallel to secant line AE. a Show that C has -coordinate r + s/. b Show that ABDE has area s r / b viewing it as a parallelogram of height s r and base of length CF. c Show that ACE has area s r /8 b observing that it has the same base and height as the parallelogram. d Compute the shaded area as the area under the graph minus the area of a trapezoid, and prove Archimedes result. B C D A r F r + s FIGURE 9 Graph of f= ab. E s a The slope of the secant line AE is fs fr s r and the slope of the tangent line along the parabola is = s ab s r ab r s r f = a + b. = a + b r + s If C is the point on the parabola at which the tangent line is parallel to the secant line AE, then its -coordinate must satisf a + b = a + b r + s or = r + s. b Parallelogram ABDE has height s r and base of length CF. Since the equation of the secant line AE is the length of the segment CF is r + s = [a + b r + s] r + r ab r, a b r + s r + s [a + b r + s] r r ab r = Thus, the area of ABDE is s r. c Triangle ACE is comprised of ACF and CEF. Each of these smaller triangles has height s r s r. Thus, the area of ACE is s r s r + s r s r = d The area under the graph of the parabola between = r and = s is s ab d = ab + a + b s r r s r. 8 = abs + a + bs s + abr a + br + r s r. and base of length = abr s + a + bs rs + r + r sr + rs + s, April,

57 SECTIO 5. The Fundamental Theorem of Calculus, Part I 6 while the area of the trapezoid under the shaded region is s r[s ab s + r ab r] = [ s r ab + a + br + s r s ] Thus, the area of the shaded region is r s r + rs + s r s which is four-thirds the area of the triangle ACE. = abr s + a + bs rr + s + r sr + s. = s r 6 r rs + 6 s = 6 s r, 6. a Appl the Comparison Theorem Theorem 5 in Section 5. to the inequalit sin valid for to prove that b Appl it again to prove that cos sin for 6 c Verif these inequalities for =.. a We have sin tdt= cos t = cos + and tdt= t =. Hence cos +. Solving, this gives cos. cos follows automaticall. b The previous part gives us t [,], sin. 6 c Substituting =. into the inequalities obtained in a and b ields respectivel. cos t, for t>. Theorem 5 gives us, after integrating over the interval and , 65. Use the method of Eercise 6 to prove that 6 cos + sin for Verif these inequalities for =.. Wh have we specified for sin but not for cos? B Eercise 6, t 6 t sin t t for t>. Integrating this inequalit over the interval [,], and then solving for cos, ields: cos cos +. These inequalities appl for. Since cos,, and + are all even functions, the also appl for. April,

58 6 CHAPTER 5 THE ITEGRAL Having established that t cos t t + t, for all t, we integrate over the interval [,], to obtain: 6 sin The functions sin, 6 and are all odd functions, so the inequalities are reversed for <. Evaluating these inequalities at =. ields , both of which are true. 66. Calculate the net pair of inequalities for sin and cos b integrating the results of Eercise 65. Can ou guess the general pattern? Integrating over the interval [,] ields Solving for cos and ields t t 6 sin t t t 6 + t5 for t cos cos 7 +. Replacing each b t and integrating over the interval [,] produces sin To see the pattern, it is best to compare consecutive inequalities for sin and those for cos : sin 6 sin 6 sin Each iteration adds an additional term. Looking at the highest order terms, we get the following pattern: 6 =! 5 5! We guess that the leading term of the polnomials are of the form n n+ n +!. Similarl, for cos, the leading terms of the polnomials in the inequalit are of the form n n n!. 67. Use FTC I to prove that if f K for [a,b], then f fa K a for [a,b]. Let a>bbe real numbers, and let fbe such that f K for [a,b]. B FTC, f t dt = f fa. a April,

59 SECTIO 5. The Fundamental Theorem of Calculus, Part II 6 Since f K for all [a,b], we get: f fa= f t dt K a. a Since f K for all [a,b], we get: f fa= f t dt K a. a Combining these two inequalities ields so that, b definition, K a f fa K a, f fa K a. 68. a Use Eercise 67 to prove that sin a sin b a b for all a,b. b Let f= sin + a sin. Use part a to show that the graph of flies between the horizontal lines =±a. c Plot fand the lines =±a to verif b for a =.5 and a =.. a Let f= sin, so that f = cos, and for all. From Eercise 67, we get: f sin a sin b a b. b Let f= sin + a sin. Appling a, we get the inequalit: This is equivalent, b definition, to the two inequalities: f = sin + a sin + a = a. a sin + a sin a. c The plots of = sin +.5 sin and of = sin +. sin are shown below. The inequalit is satisfied in both plots The Fundamental Theorem of Calculus, Part II Preliminar Questions. Let G = t + dt. a Is the FTC needed to calculate G? b Is the FTC needed to calculate G? a o. G = t + dt =. b Yes. B the FTC II, G = +, so G = 65.. Which of the following is an antiderivative Fof f= satisfing F =? a t dt b t dt c The correct answer is c: t dt. t dt April,

60 6 CHAPTER 5 THE ITEGRAL. Does ever continuous function have an antiderivative? Eplain. Yes. All continuous functions have an antiderivative, namel ftdt. a. Let G = sin tdt. Which of the following statements are correct? a G is the composite function sin. b G is the composite function A, where A = sint dt c G is too complicated to differentiate. d The Product Rule is used to differentiate G. e The Chain Rule is used to differentiate G. f G = sin. Statements b, e, and f are correct. Eercises. Write the area function of f= + with lower limit a = as an integral and find a formula for it. Let f= +. The area function with lower limit a = is A = ftdt = t + dt. a Carring out the integration, we find t + dt = t + t = + + = + + or +. Therefore, A = +.. Find a formula for the area function of f= + with lower limit a =. The area function for f= + with lower limit a = is given b A = t + dt = t + t = +.. Let G = t dt. Calculate G, G and G. Then find a formula for G. Let G = t dt. Then G = t dt =. Moreover, G =, so that G = and G =. Finall, G = t dt = t t = = Find F, F, and F, where F = t + tdt. B definition, F = t + tdt =. B FTC, F = +, so that F = + = and F = + = =. 5. Find G, G, and G π/, where G = tan tdt. B definition, G = tan tdt =. B FTC, G = tan, so that G = tan = and G π = tan π =. 6. Find H and H du, where H = u +. du B definition, H = u + =. B FTC, H = +,soh = 5. In Eercises 7 6, find formulas for the functions represented b the integrals. 7. u du F = u du = 5 u5 = April,

61 SECTIO 5. The Fundamental Theorem of Calculus, Part II t 8tdt F = t 8tdt = t t = sin udu F = sin udu= cos u = cos.. sec θdθ π/ F = sec θdθ= tan θ = tan tan π/ = tan +. π/ π/. e u du F = e u du = eu = e e.. e t dt F = e t dt = e t = + e.. tdt F = tdt= t =. /. sec udu / / F = sec / udu= tan u = tan / / tan e u du 9+ F = e u du = e u 9+ = e 9 + e. dt 6. t dt = ln t t = ln ln = ln ln. In Eercises 7, epress the antiderivative Fof fsatisfing the given initial condition as an integral. 7. f= +, F5 = The antiderivative Fof + satisfing F5 = isf = 5 8. f= + + 9, F7 = The antiderivative Fof f= + t + dt. satisfing F7 = isf = 7 t + t + 9 dt f= sec, F = The antiderivative F of f = sec satisfing F = isf = sec tdt.. f= e, F = The antiderivative Fof f= e satisfing F = is F = e t dt. April,

62 66 CHAPTER 5 THE ITEGRAL In Eercises, calculate the derivative. d. t 5 9t dt d d B FTC II, t 5 9t dt = 5 9. d d θ. cot udu dθ B FTC II, d t sec5 9d dt d d θ cot udu= cot θ. dθ. t B FTC II, sec5 9d = sec5t 9. dt d s. tan ds + u du d s B FTC II, tan ds + u du = tan + s. 5. Let A = ftdt for fin Figure 8. a Calculate A, A, A, and A. b Find formulas for A on [, ] and [, ] and sketch the graph of A. = f FIGURE 8 a A = =, the area under ffrom = to =, while A = + = 6.5, the area under f from = to =. B the FTC, A = fso A = f = and A = f =. b For each [, ], the region under the graph of = fis a rectangle of length and height ; for each [, ], the region is comprised of a square of side length and a trapezoid of height and bases and. Hence, {, < A = +, A graph of the area function A is shown below. 6. Make a rough sketch of the graph of A = 8 6 Area Function A gt dt for g in Figure 9. = g FIGURE 9 April,

63 SECTIO 5. The Fundamental Theorem of Calculus, Part II 67 The graph of = g lies above the -ais over the interval [, ], below the -ais over [, ], and above the -ais over [, ]. The corresponding area function should therefore be increasing on,, decreasing on, and increasing on,. Further, it appears from Figure 9 that the local minimum of the area function at = should be negative. One possible graph of the area function is the following. 7. Verif: Let ft= t = t dt =. Hint: Consider and separatel. { t for t t for t<. Then tdt for t = for F = ftdt = = t dt for < t = for < For, we have F = = since =, while for <, we have F = = since =. Therefore, for all real we have F =. 8. Find G, where G = t + dt. B combining the Chain Rule and FTC, G = 6 +, sog = + =. In Eercises 9, calculate the derivative. 9. d tdt d t +. d d. d ds B the Chain Rule and the FTC, / cos tdt B the Chain Rule and the FTC, cos s u du 6 B the Chain Rule and the FTC, d. tdt d Hint for Eercise : F = A A. Let d d d d d ds tdt t + = + = / cos tdt= cos +. = cos cos s u du = cos s sin s = cos s sin s. 6 F = tdt= tdt tdt. Appling the Chain Rule combined with FTC, we have F = = 5.. April,

64 68 CHAPTER 5 THE ITEGRAL. d d tan tdt Let G = tan tdt= tan tdt tan tdt. Appling the Chain Rule combined with FTC twice, we have G = tan tan / = tan tan. u d. du u + d Let u u u G = + d = + d + d. u Appling the Chain Rule combined with FTC twice, we have G = 9u + + u +. In Eercises 5 8, with fas in Figure let A = ftdt and B = ftdt. = f 5 6 FIGURE 5. Find the min and ma of A on [, 6]. The minimum values of A on [, 6] occur where A = f goes from negative to positive. This occurs at one place, where =.5. The minimum value of A is therefore A.5 =.5. The maimum values of A on [, 6] occur where A = fgoes from positive to negative. This occurs at one place, where =.5. The maimum value of A is therefore A.5 = Find the min and ma of B on [, 6]. The minimum values of B on [, 6] occur where B = f goes from negative to positive. This occurs at one place, where =.5. The minimum value of A is therefore B.5 =.5. The maimum values of B on [, 6] occur where B = fgoes from positive to negative. This occurs at one place, where =.5. The maimum value of B is therefore B.5 = Find formulas for A and B valid on [, ]. On the interval [, ], A = B = f=. A = ftdt = and B = ftdt =. Hence A = and B =. 8. Find formulas for A and B valid on [, 5]. On the interval [, 5], A = B = f =.5 = 9. A = ftdt = and B = ftdt =. Hence A = 9 9 and B = Let A = ftdt, with fas in Figure. a Does A have a local maimum at P? b Where does A have a local minimum? April,

65 SECTIO 5. The Fundamental Theorem of Calculus, Part II 69 c Where does A have a local maimum? d True or false? A < for all in the interval shown. P = f Q R S FIGURE Graph of f. a In order for A to have a local maimum, A = fmust transition from positive to negative. As this does not happen at P, A does not have a local maimum at P. b A will have a local minimum when A = f transitions from negative to positive. This happens at R, so A has a local minimum at R. c A will have a local maimum when A = f transitions from positive to negative. This happens at S, so A has a local maimum at S. d It is true that A < oni since the signed area from to is clearl alwas negative from the figure.. Determine f, assuming that ftdt = +. Let F = ftdt = +. Then F = f= +.. Determine the function g and all values of c such that gt dt = + 6 c B the FTC II we have and therefore, g = d d + 6 = + gt dt = + c + c c We must choose c so that c + c = 6. We can take c = orc =. b. Find a b such that 9d has minimal value. a Let a be given, and let F a = a t 9dt. Then F a = 9, and the critical points are =±. Because F a = 6and F a = 6, we see that F a has a minimum at =. ow, we find a minimizing a 9d. Let G = 9d. Then G = 9, ielding critical points = or =. With =, G = 9d = 9 = 6. With =, G = 9d =. b Hence a = and b = are the values minimizing 9d. a In Eercises and, let A = ftdt. a. Area Functions and Concavit Eplain wh the following statements are true. Assume fis differentiable. a If c is an inflection point of A, then f c =. b A is concave up if fis increasing. c A is concave down if fis decreasing. April,

66 6 CHAPTER 5 THE ITEGRAL a If = c is an inflection point of A, then A c = f c =. b If A is concave up, then A >. Since A is the area function associated with f, A = fb FTC II, so A = f. Therefore f >, so fis increasing. c If A is concave down, then A <. Since A is the area function associated with f, A = fb FTC II, so A = f. Therefore, f < and so fis decreasing.. Match the propert of A with the corresponding propert of the graph of f. Assume fis differentiable. Area function A a A is decreasing. b A has a local maimum. c A is concave up. d A goes from concave up to concave down. Graph of f i Lies below the -ais. ii Crosses the -ais from positive to negative. iii Has a local maimum. iv fis increasing. Let A = a ftdt be an area function of f. Then A = fand A = f. a A is decreasing when A = f<, i.e., when flies below the -ais. This is choice i. b A has a local maimum at when A = fchanges sign from + toto as increases through, i.e., when fcrosses the -ais from positive to negative. This is choice ii. c A is concave up when A = f >, i.e., when fis increasing. This corresponds to choice iv. d A goes from concave up to concave down at when A = f changes sign from + toto as increases through, i.e., when fhas a local maimum at. This is choice iii. 5. Let A = ftdt, with fas in Figure. Determine: a The intervals on which A is increasing and decreasing b The values where A has a local min or ma c The inflection points of A d The intervals where A is concave up or concave down = f 6 8 FIGURE a A is increasing when A = f>, which corresponds to the intervals, and 8,. A is decreasing when A = f<, which corresponds to the intervals, 8 and,. b A has a local minimum when A = f changes from to +, corresponding to = 8. A has a local maimum when A = fchanges from + to, corresponding to = and =. c Inflection points of A occur where A = f changes sign, or where f changes from increasing to decreasing or vice versa. Consequentl, A has inflection points at =, = 6, and =. d A is concave up when A = f is positive or fis increasing, which corresponds to the intervals, and 6,. Similarl, A is concave down when f is decreasing, which corresponds to the intervals, 6 and,. 6. Let f= 5 6 and F = ftdt. a Find the critical points of Fand determine whether the are local minima or local maima. b Find the points of inflection of Fand determine whether the concavit changes from up to down or from down to up. c Plot fand Fon the same set of aes and confirm our answers to a and b. April,

67 SECTIO 5. The Fundamental Theorem of Calculus, Part II 6 a If F = t 5t 6dt, then F = 5 6 and F = 5. Solving F = 5 6 = ields critical points = and = 6. Since F = 7 <, there is a local maimum value of F at =. Moreover, since F 6 = 7 >, there is a local minimum value of F at = 6. b As noted in part a, F = 5 6 and F = 5. A candidate point of inflection occurs where F = 5 =. Thus = 5. F changes from negative to positive at this point, so there is a point of inflection at = 5 and concavit changes from down to up. c From the graph below, we clearl note that Fhas a local maimum at =, a local minimum at = 6 and a point of inflection at = 5. 6 f F 7. Sketch the graph of an increasing function fsuch that both f and A = ftdt are decreasing. If f is decreasing, then f must be negative. Furthermore, if A = ftdt is decreasing, then A = fmust also be negative. Thus, we need a function which is negative but increasing and concave down. The graph of one such function is shown below. 8. Figure shows the graph of f= sin. Let F = t sin tdt. a Locate the local ma and absolute ma of Fon [, π]. b Justif graphicall: Fhas precisel one zero in [π, π]. c How man zeros does Fhave in [, π]? d Find the inflection points of Fon [, π]. For each one, state whether the concavit changes from up to down or from down to up. 8 p p p p 5p p FIGURE Graph of f= sin. Let F = t sin tdt. A graph of f = sin is depicted in Figure. ote that F = fand F = f. a For F to have a local maimum at, π we must have F = f = and F = f must change sign from + toto as increases through. This occurs at = π. The absolute maimum of Fon [, π] occurs at = π since from the figure the signed area between = and = c is greatest for = c = π. b At = π, the value of F is positive since f>on,π.as increases along the interval [π, π], we see that F decreases as the negativel signed area accumulates. Eventuall the additional negativel signed area outweighs the prior positivel signed area and F attains the value, sa at b π, π. Thereafter, on b, π, we see that f is negative and thus F becomes and continues to be negative as the negativel signed area accumulates. Therefore, F takes the value eactl once in the interval [π, π]. April,

68 6 CHAPTER 5 THE ITEGRAL c Fhas two zeroes in [, π]. One is described in part b and the other must occur in the interval [π, π] because F < at = π but clearl the positivel signed area over [π, π] is greater than the previous negativel signed area. d Since f is differentiable, we have that F is twice differentiable on I. Thus Fhas an inflection point at provided F = f = and F = f changes sign at.iff = f changes sign from + toto at, then f has a local maimum at. There is clearl such a value in the figure in the interval [π/,π] and another around 5π/. Accordingl, F has two inflection points where F changes from concave up to concave down. If F = f changes sign from toto+ at, then f has a local minimum at. From the figure, there is such an around π/; so F has one inflection point where Fchanges from concave down to concave up. 9. Find the smallest positive critical point of F = cost / dt and determine whether it is a local min or ma. Then find the smallest positive inflection point of F and use a graph of = cos / to determine whether the concavit changes from up to down or from down to up. A critical point of F occurs where F = cos / =. The smallest positive critical points occurs where / = π/, so that = π/ /. F goes from positive to negative at this point, so = π/ / corresponds to a local maimum.. Candidate inflection points of F occur where F =. B FTC, F = cos /, so F = / / sin /. Finding the smallest positive of F =, we get: / / sin / = sin / = / = π since > = π /.5. From the plot below, we see that F = cos / changes from decreasing to increasing at π /,sof changes from concave down to concave up at that point..5.5 Further Insights and Challenges 5. Proof of FTC II The proof in the tet assumes that fis increasing. To prove it for all continuous functions, let mh and Mh denote the minimum and maimum of fton [, + h] Figure. The continuit of fimplies that lim mh = lim Mh = f. Show that for h>, h h hmh A + h A hmh For h<, the inequalities are reversed. Prove that A = f. = f Mh mh a + h FIGURE Graphical interpretation of A + h A. Let f be continuous on [a,b]. For h>, let mh and Mh denote the minimum and maimum values of f on [, + h]. Since f is continuous, we have lim mh = lim Mh = f.ifh>, then since h + h + mh f Mh on [, + h], we have +h +h +h +h hmh = mh dt ftdt = A + h A = ftdt Mh dt = hmh. April,

69 SECTIO 5. The Fundamental Theorem of Calculus, Part II 6 In other words, hmh A + h A hmh. Since h>, it follows that mh Letting h + ields whence f lim h + A + h A h f, A + h A lim = f h + h A + h A h b the Squeeze Theorem. If h<, then hmh = mh dt ftdt = A A + h = ftdt Mh dt = hmh. +h +h +h +h Since h<, we have h > and thus or Letting h gives mh mh f lim h A A + h h A + h A h A + h A h Mh Mh. f, so that A + h A lim = f h h b the Squeeze Theorem. Since the one-sided limits agree, we therefore have A A + h A = lim = f. h h Mh. 5. Proof of FTC I FTC I asserts that b a ftdt = Fb Faif F = f. Use FTC II to give a new proof of FTC I as follows. Set A = a ftdt. a Show that F = A + C for some constant. b b Show that Fb Fa = Ab Aa = ftdt. a Let F = fand A = a ftdt. a Then b the FTC, Part II, A = fand thus A and F are both antiderivatives of f. Hence F = A + C for some constant C. b Fb Fa = Ab + C Aa + C = Ab Aa b a b b = ftdt ftdt = ftdt = ftdt a a a a which proves the FTC, Part I. 5. Can EverAntiderivative Be Epressed as an Integral? The area function a ftdtis an antiderivative of f for ever value of a. However, not all antiderivatives are obtained in this wa. The general antiderivative of f= is F = + C. Show that Fis an area function if C but not if C>. Let f=. The general antiderivative of fis F = + C. Let A = a ftdt = a tdt= t = a a be an area function of f =. To epress F as an area function, we must find a value for a such that a = + C, whence a =± C.IfC, then C and we ma choose either a = C or a = C. However, if C>, then there is no real for a and Fcannot be epressed as an area function. 5. Prove the formula d v ftdt = f vv f uu d u April,

70 6 CHAPTER 5 THE ITEGRAL Write v v v u fd = fd+ fd = fd fd. u u Then, b the Chain Rule and the FTC, d v fd = d v fd d u fd d u d d = f vv f uu. 5. Use the result of Eercise 5 to calculate B Eercise 5, d e sin tdt d ln d e sin tdt= e sin e sin ln. d ln 5.5 et Change as the Integral of a Rate Preliminar Questions. A hot metal object is submerged in cold water. The rate at which the object cools in degrees per minute is a function ftof time. Which quantit is represented b the integral T ftdt? The definite integral T ftdt represents the total drop in temperature of the metal object in the first T minutes after being submerged in the cold water.. A plane travels 56 km from Los Angeles to San Francisco in hour. If the plane s velocit at time t is vt km/h, what is the value of vt dt? The definite integral vt dt represents the total distance traveled b the airplane during the one hour flight from Los Angeles to San Francisco. Therefore the value of vt dt is 56 km.. Which of the following quantities would be naturall represented as derivatives and which as integrals? a Velocit of a train b Rainfall during a 6-month period c Mileage per gallon of an automobile d Increase in the U.S. population from 99 to Quantities a and c involve rates of change, so these would naturall be represented as derivatives. Quantities b and d involve an accumulation, so these would naturall be represented as integrals. Eercises. Water flows into an empt reservoir at a rate of + t liters per hour. What is the quantit of water in the reservoir after 5 hours? The quantit of water in the reservoir after five hours is 5 + tdt = t + t 5 = 5,5 gallons.. A population of insects increases at a rate of + t +.5t insects per da. Find the insect population after das, assuming that there are 5 insects at t =. The increase in the insect population over three das is + t + t dt = t + 5t + t = 589 = Accordingl, the population after das is = 68.5 or 68 insects. April,

71 SECTIO 5.5 et Change as the Integral of a Rate 65. A surve shows that a maoral candidate is gaining votes at a rate of t + votes per da, where t is the number of das since she announced her candidac. How man supporters will the candidate have after 6 das, assuming that she had no supporters at t =? The number of supporters the candidate has after 6 das is 6 t + dt = t 6 + t =,66,.. A factor produces biccles at a rate of 95 + t t biccles per week. How man biccles were produced from the beginning of week to the end of week? The rate of production is rt = 95 + t t biccles per week and the period from the beginning of week to the end of week corresponds to the second and third weeks of production. Accordingl, the number of bikes produced from the beginning of week to the end of week is rtdt = 95 + t t dt = 95t + t t = biccles. 5. Find the displacement of a particle moving in a straight line with velocit vt = t m/s over the time interval [, 5]. The displacement is given b 5 t dt = t 5 t = = m. 6. Find the displacement over the time interval [, 6] of a helicopter whose vertical velocit at time t is vt =.t + t m/s. Given vt = 5 t + t m/s, the change in height over [, 6] is 6 6 vt dt = 5 t + t dt = 5 t + 6 t = m A cat falls from a tree with zero initial velocit at time t =. How far does the cat fall between t =.5 and t = s? Use Galileo s formula vt = 9.8t m/s. Given vt = 9.8t m/s, the total distance the cat falls during the interval [, ] is / vt dt = 9.8t dt=.9t =.9.5 =.675 m. / / 8. A projectile is released with an initial vertical velocit of m/s. Use the formula vt = 9.8t for velocit to determine the distance traveled during the first 5 seconds. The distance traveled is given b 5 / t dt = 9.8tdt + 9.8t dt /9.8 = t.9t / t 5 t 6.9m. /9.8 In Eercises 9, a particle moves in a straight line with the given velocit in m/s. Find the displacement and distance traveled over the time interval, and draw a motion diagram like Figure with distance and time labels. 9. vt = t, [, 5] 5 Displacement is given b tdt = t t 5 = ft, while total distance is given b 5 5 t dt = tdt + t dt = t t + t 5 t = 6 ft. The displacement diagram is given here. April,

72 66 CHAPTER 5 THE ITEGRAL t = 5 t = t = Distance 8. vt = 6 t + t, [, ] Let vt = 6 t + t = t t 6. Displacement is given b 6 t + t dt = 6t t + t = 6 meters. Total distance traveled is given b 6 6 t + t dt = 6 t + t dt + t 6 t dt + 6 t + t dt 6 = 6t t + t + t 6t t + 6t t + t 6 = meters. The displacement diagram is given here. t = t = 6 t = vt = t, [.5, ] Displacement is given b t dt = t t = m, while total distance is given b.5.5 t dt = t dt + t dt = t t + t + t = m The displacement diagram is given here. t = t = t = Distance.5. vt = cos t, [, π] π π Displacement is given b cos tdt= sin t = meters, while the total distance traveled is given b π π/ π/ 5π/ π cos t dt = cos tdt cos tdt+ cos tdt cos t,dt π/ π/ 5π/ π/ π/ 5π/ π = sin t sin t + sin t sin t π/ π/ 5π/ = 6 meters. The displacement diagram is given here. t = π..5 t = π t = 5π π t =.5. April,

73 SECTIO 5.5 et Change as the Integral of a Rate 67. Find the net change in velocit over [, ] of an object with at = 8t t m/s. The net change in velocit is atdt = 8t t dt = t t = 9 m/s.. Show that if acceleration is constant, then the change in velocit is proportional to the length of the time interval. Let at = a be the constant acceleration. Let vt be the velocit. Let [t,t ] be the time interval concerned. We know that v t = a, and, b FTC, t vt vt = adt = at t, t So the net change in velocit is proportional to the length of the time interval with constant of proportionalit a. 5. The traffic flow rate past a certain point on a highwa is qt = + t t t in hours, where t = is 8 am. How man cars pass b in the time interval from 8 to am? The number of cars is given b qtdt = + t t dt = t + t t = + 8 = 9 cars. 6. The marginal cost of producing tablet computers is C =.6 +. What is the cost of producing units if the setup cost is $9,? If production is set at units, what is the cost of producing additional units? The production coot for producing units is.6 +. d =. +. = 6, 7, + 9, = 8, dollars. Adding in the setup cost, we find the total cost of producing units is $7,. If production is set at units, the cost of producing an additional units is.6 +. d =. +. = 8, 7, + 9,6.67 8, or $ A small boutique produces wool sweaters at a marginal cost of 5[[/5]] for, where [[]] is the greatest integer function. Find the cost of producing sweaters. Then compute the average cost of the first sweaters and the last sweaters. The total cost of producing sweaters is 5 5 5[[/5]] d = d + 5 d + d + 5 d 5 5 = = 65 dollars. From this calculation, we see that the cost of the first sweaters is $75 and the cost of the last ten sweaters is $75; thus, the average cost of the first ten sweaters is $7.5 and the average cost of the last ten sweaters is $ The rate in liters per minute at which water drains from a tank is recorded at half-minute intervals. Compute the average of the left- and right-endpoint approimations to estimate the total amount of water drained during the first minutes. t min rl/min April,

74 68 CHAPTER 5 THE ITEGRAL Let t =.5. Then R = = 9. liters L = = 5. liters The average of R and L is = liters. 9. The velocit of a car is recorded at half-second intervals in feet per second. Use the average of the left- and right-endpoint approimations to estimate the total distance traveled during the first seconds. t vt Let t =.5. Then The average of R and L is.5 ft. R = = ft. L = = 5 ft.. To model the effects of a carbon ta on CO emissions, policmakers stud the marginal cost of abatement B, defined as the cost of increasing CO reduction from to + tons in units of ten thousand tons Figure. Which quantit is represented b the area under the curve over [, ] in Figure? B $/ton Tons reduced in ten thousands FIGURE Marginal cost of abatement B. The area under the curve over [, ] represents the total cost of reducing the amount of CO released into the atmosphere b tons.. A megawatt of power is 6 W, or.6 9 J/hour. Which quantit is represented b the area under the graph in Figure 5? Estimate the energ in joules consumed during the period pm to 8 pm. Megawatts in thousands Hour of the da FIGURE 5 Power consumption over -da period in California Februar. The area under the graph in Figure 5 represents the total power consumption over one da in California. Assuming t = corresponds to midnight, the period pm to 8 pm corresponds to t = 6 to t =. The left and right endpoint approimations are The average of these values is L = = 99.megawatt hours R = =.megawatt hours.75megawatt hours =.67 joules. April,

75 SECTIO 5.5 et Change as the Integral of a Rate 69. Figure 6 shows the migration rate Mt of Ireland in the period This is the rate at which people in thousands per ear move into or out of the countr. a Is the following integral positive or negative? What does this quantit represent? 998 Mtdt 988 b Did migration in the period result in a net influ of people into Ireland or a net outflow of people from Ireland? c During which two ears could the Irish prime minister announce, We ve hit an inflection point. We are still losing population, but the trend is now improving. Migration in thousands FIGURE 6 Irish migration rate in thousands per ear. a Because there appears to be more area below the t-ais than above in Figure 6, 998 Mtdt 988 is negative. This quantit represents the net migration from Ireland during the period b As noted in part a, there appears to be more area below the t-ais than above in Figure 6, so migration in the period resulted in a net outflow of people from Ireland. c The prime minister can make this statement when the graph of M is at a local minimum, which appears to be in the ears 989 and 99.. Let d be the number of asteroids of diameter d kilometers. Data suggest that the diameters are distributed according to a piecewise power law: {.9 9 d. for d<7 d =.6 d for d 7 a Compute the number of asteroids with diameter between. and km. b Using the approimation d + d d, estimate the number of asteroids of diameter 5 km. a The number of asteroids with diameter between. and km is 7 d dd =.9 9 d. dd +.6 d dd.. 7 b Taking d = 9.5, = d....6 d 7 = = =, Thus, there are approimatel,56 asteroids of diameter 5 km.. Heat Capacit The heat capacit CT of a substance is the amount of energ in joules required to raise the temperature of gb C at temperature T. a Eplain wh the energ required to raise the temperature from T to T is the area under the graph of CT over [T,T ]. b How much energ is required to raise the temperature from 5 to CifCT = 6 +. T? April,

76 6 CHAPTER 5 THE ITEGRAL a Since CT is the energ required to raise the temperature of one gram of a substance b one degree when its temperature is T, the total energ required to raise the temperature from T to T is given b the definite integral T T CT dt.asct >, the definite integral also represents the area under the graph of CT. b If CT = 6 +. T = T /, then the energ required to raise the temperature from 5 Cto Cis 5 CT dt or T / dt = 6T + 5 T / = / / = 86.9 Joules 5. Figure 7 shows the rate Rt of natural gas consumption in billions of cubic feet per da in the mid-atlantic states ew York, ew Jerse, Pennslvania. Epress the total quantit of natural gas consumed in 9 as an integral with respect to time t in das. Then estimate this quantit, given the following monthl values of Rt:.8,.86,.9,.9,.8,.8,.,.89,.89,.,.6,.5 Keep in mind that the number of das in a month varies with the month. atural gas consumption 9 cubic ft/da J F M A M J J A S O D FIGURE 7 atural gas consumption in 9 in the mid-atlantic states The total quantit of natural gas consumed is given b 65 Rt dt. With the given data, we find 65 Rt dt = 65.5 billion cubic feet. 6. Cardiac output is the rate R of volume of blood pumped b the heart per unit time in liters per minute. Doctors measure R b injecting A mg of de into a vein leading into the heart at t = and recording the concentration ct of de in milligrams per liter pumped out at short regular time intervals Figure 8. a Eplain: The quantit of de pumped out in a small time interval [t,t + t] is approimatel Rct t. b Show that A = R T ct dt, where T is large enough that all of the de is pumped through the heart but not so large that the de returns b recirculation. c Assume A = 5 mg. Estimate R using the following values of ct recorded at -second intervals from t = to t = :,.,.8, 6.5, 9.8, 8.9, 6.,,.,., Inject de here Measure concentration here ct mg/l Blood flow = ct t s FIGURE 8 April,

77 SECTIO 5.5 et Change as the Integral of a Rate 6 a Over a short time interval, ct is nearl constant. Rct is the rate of volume of de amount of fluid concentration of de in fluid flowing out of the heart in mg per minute. Over the short time interval [t,t + t], the rate of flow of de is approimatel constant at Rct mg/minute. Therefore, the flow of de over the interval is approimatel Rct t mg. b The rate of flow of de is Rct. Therefore the net flow between time t = and time t = T is T T Rct dt = R ct dt. If T is great enough that all of the de is pumped through the heart, the net flow is equal to all of the de, so T A = R ct dt. T c In the table, t = 6 minute, and =. The right and left hand approimations of ct dt are: R = mg minute = liter L = mg minute = liter Both L and R are the same, so the average of L and R is.698. Hence, T A = R ctdt mg minute 5mg= R.698 liter R = liters minute = 7.6 liters minute. Eercises 7 and 8: A stud suggests that the etinction rate rt of marine animal families during the Phanerozoic Eon can be modeled b the function rt = /t + 6 for t 5, where t is time elapsed in millions of ears since the beginning of the eon 5 million ears ago. Thus, t = 5 refers to the present time, t = 5 is million ears ago, and so on. 7. Compute the average of R and L with = 5 to estimate the total number of families that became etinct in the periods t 5 and 5 t. t 5 For = 5, t = The table of values {rt i } i=...5 is given below: The endpoint approimations are: 5 5 =. t i 5 rt i R = families L = families The right endpoint approimation estimates families became etinct in the period t 5, the left endpoint approimation estimates.9 families became etinct during this time. The average of the two is 5.6 families. 5 t For =, t = 5 5 = 9. April,

78 6 CHAPTER 5 THE ITEGRAL The table of values {rt i } i=...5 is given below: The endpoint approimations are: t i rt i R = families L = families The right endpoint approimation estimates.888 families became etinct in the period 5 t, the left endpoint approimation estimates 7.75 families became etinct during this time. The average of the two is 5.8 families. 8. Estimate the total number of etinct families from t = to the present, using M with = 5. We are estimating 5 t + 6 dt using M with = 5. If = 5, t = 5 5 M = t i= = and {t i } i=,... = i t t/ = i. 5 rti = i = 57.. i= Thus, we estimate that 57 families have become etinct over the past 5 million ears. Further Insights and Challenges 9. Show that a particle, located at the origin at t = and moving along the -ais with velocit vt = t, will never pass the point =. The particle s velocit is vt = s t = t, an antiderivative for which is Ft = t. Hence, the particle s position at time t is t st = s t u du = F u = Ft F = < t for all t. Thus, the particle will never pass =, which implies it will never pass = either.. Show that a particle, located at the origin at t = and moving along the -ais with velocit vt = t / moves arbitraril far from the origin after sufficient time has elapsed. The particle s velocit is vt = s t = t /, an antiderivative for which is Ft = t /. Hence, the particle s position at time t is t st = s t u du = F u = Ft F = t for all t. Let S>denote an arbitraril large distance from the origin. We see that for S + t>, the particle will be more than S units from the origin. In other words, the particle moves arbitraril far from the origin after sufficient time has elapsed. 5.6 Substitution Method Preliminar Questions. Which of the following integrals is a candidate for the Substitution Method? a 5 sin 5 d b sin 5 cos d c 5 sin d April,

79 SECTIO 5.6 Substitution Method 6 The function in c: 5 sin is not of the form guu. The function in a meets the prescribed pattern with gu = sin u and u = 5. Similarl, the function in b meets the prescribed pattern with gu = u 5 and u = sin.. Find an appropriate choice of u for evaluating the following integrals b substitution: a + 9 d b sin d c sin cos d a + 9 = + 9 ; hence, c =, f u = u, and u = + 9. b sin = sin ; hence, c =, f u = sin u, and u =. c sin cos = sin cos ; hence, c =, f u = u, and u = cos.. Which of the following is equal to + d for a suitable substitution? a 9 udu b udu c 9 udu With the substitution u = +, the definite integral + d becomes 9 udu. The correct answer is c. Eercises In Eercises 6, calculate du.. u = Let u =. Then du = d.. u = + 8 Let u = + 8. Then du = 8 8 d.. u = cos Let u = cos. Then du = sin d= sin d.. u = tan Let u = tan. Then du = sec d. 5. u = e + 6. u = ln + Let u = e +. Then du = e + d. Let u = ln +. Then du = + d. In Eercises 7, write the integral in terms of u and du. Then evaluate d, u = 7 Let u = 7. Then du = d. Hence 7 d = u du = u + C = 7 + C d, u = Let u = + 5. Then du = d and + 5 d = t t + dt, u = t + u du = u + C = C. Let u = t +. Then du = t dt. Hence, t t + dt = u / du = u/ + C = t + / + C. April,

80 6 CHAPTER 5 THE ITEGRAL. + cos + d, u = + Let u = +. Then du = + d = + d and + cos + d = cos udu= sin u + C = sin + + C.. t t dt, u = t Let u = t. Then du = 8t dt. Hence, t t dt = u du = 8 8 u + C = 8 t + C.. d, u = Let u =. Then du = d or du = d. Hence u d =. + 9 d, u = + u / du = u/ + C = 6 / + C. Let u = +. Then = u and du = d. Hence + 9 d = u u 9 du = u u 9 du = u u + C = C.. d, u = Let u =. Then = u + and du = d or du = d. Hence, d = u + u / du = u / + u / du 6 6 = 6 5 u5/ + 6 u/ + C = 5/ + / + C d, u = + Let u = +. Then = u and du = d. Hence + d = u u / du = u 5/ u / + u / du = 7 u7/ 5 u5/ + u/ + C = 7 + 7/ 5 + 5/ + + / + C. 6. sinθ 7dθ, u = θ 7 Let u = θ 7. Then du = dθ and sinθ 7dθ = sin udu= cos u + C = cosθ 7 + C. April,

81 SECTIO 5.6 Substitution Method sin θ cos θdθ, u = sin θ Let u = sin θ. Then du = cos θdθ. Hence, sin θ cos θdθ= u du = u + C = sin θ + C. 8. sec tan d, u = tan Let u = tan. Then du = sec d. Hence sec tan d= udu= u + C = tan + C. 9. e d, u = Let u =. Then du = dor du = d. Hence, e d = e u du = eu + C = e + C.. sec te tan t dt, u = tan t Let u = tan t. Then du = sec tdtand sec te tan t dt = e u du = e u + C = e tan t + C.. ln d, u = ln Let u = ln. Then du = d, and ln d = u du = u + C = ln + C. tan d., u = tan + Let u = tan. Then du = + d, and tan d = u du = + u + C = tan + C. In Eercises 6, evaluate the integral in the form a sinu + C for an appropriate choice of u and constant a.. cos d Let u =. Then du = d or du = d. Hence cos d = cos udu= sin u + C = sin + C.. cos + d Let u = +. Then du = d or du = d. Hence cos + d = cos udu= sin u + C. 5. / cos / d Let u = /. Then du = / d or du = / d. Hence / cos / d = cos udu= sin u + C = sin/ + C. April,

82 66 CHAPTER 5 THE ITEGRAL 6. cos cossin d Let u = sin. Then du = cos d. Hence cos cossin d = cos udu= sin u + C. In Eercises 7 7, evaluate the indefinite integral d Let u = + 5. Then du = d and d = u 9 du = u + C = C. d 9 5 Let u = 9. Then du = d and d 9 5 = u 5 du = u + C = 9 + C. dt t + Let u = t +. Then du = dt and dt = u / du = u / + C = t + + C. t +. 9t + / dt. Let u = 9t +. Then du = 9 dt and 9t + / dt = u / du = u5/ + C = 5 9t + 5/ + C. + + d Let u = +. Then du = + d or du = + d. Hence + + d = u du = u + C = + + C =. + + / d + + C. Let u = +. Then du = + d = + d and + + / d = u / du = 7 u7/ + C. + 9 d = 7 + 7/ + C. Let u = + 9. Then du = dor du = d. Hence + 9 d = du = u + C = u C. April,

83 . + + d SECTIO 5.6 Substitution Method 67 Let u = +. Then du = + 6d or 6 du = + d. Hence + + d = u du = 6 6 u + C = C d Let u = +. Then du = + d. Hence + + d = u du = u + C = + + C d Let u = 5 +. Then du = 5 + d. Hence d = u du = u + C = C d Let u = + 8. Then du = d and + 8 d = u du = 6 u + C = C d Let u = + 8. Then du = d, = u 8, and + 8 d = u 8u du = u 8u du 9 9 = 9 u u + C 9. + d = C. Let u = +. Then du = d and + d = u / du = 9 u/ + C = 9 + / + C d Let u = +. Then du = d, = u and 5 + d = u udu= u / u / du = 5 u5/ u/ + C. d + 5 = 5 + 5/ 9 + / + C. Let u = + 5. Then du = d and d + 5 = u du = u + C = C. April,

84 68 CHAPTER 5 THE ITEGRAL. d + 5 Let u = + 5. Then du = d, = u 5 and d u = u du = u u + 5u du = ln u +u 5 u + C = ln C.. z z + dz Let u = z +. Then du = z dz and z z + dz = u du = 9 u + C = 9 z + + C.. z 5 + z z + dz Let u = z +. Then du = z dz, z = u and z 5 + z z + dz = u + u du = u + u du = u + u + C / d = z + + z + + C. Let u = +. Then = u, du = d and + + / d = u + u / du = u 5/ + u / du = 9 u9/ + 5 u5/ + C = 9 + 9/ / + C. 6. / d Let u =. Then u + = and du = dor du = d. Hence / d = / d 7. sin8 θdθ = u + u / du = u 5/ + u / du = 7 u7/ + 5 u5/ + C = 7 7/ + 5 5/ + C. Let u = 8 θ. Then du = dθ and sin8 θdθ = sin udu= cos u + C = cos8 θ+ C. April,

85 8. θ sinθ dθ SECTIO 5.6 Substitution Method 69 Let u = θ. Then du = θ dθand θ sinθ dθ = sin udu= cos u + C = cosθ + C. 9. cos t t dt Let u = t = t /. Then du = t / dt and cos t dt = cos udu= sin u + C = sin t + C. t 5. sin + d Let u = +. Then du = d or du = d. Hence sin + d = sin udu= cos u + C = cos + + C. 5. tanθ + 9dθ Let u = θ + 9. Then du = dθ and tanθ + 9dθ = tan udu= ln sec u +C = ln secθ + 9 +C. 5. sin 8 θ cos θdθ Let u = sin θ. Then du = cos θdθand sin 8 θ cos θdθ= u 8 du = 9 u9 + C = 9 sin9 θ + C. 5. cot d Let u = sin. Then du = cos d, and cos cot d= sin d = 5. /5 tan /5 d du u = ln u +C = ln sin +C. Let u = /5. Then du = 5 /5 d and /5 tan /5 d = 5 tan udu= 5 ln sec u +C = 5 ln sec /5 +C. 55. sec + 9d Let u = + 9. Then du = d or du = d. Hence sec + 9d = sec udu= tan u + C = tan C. 56. sec tan d Let u = tan. Then du = sec d. Hence sec tan d= u du = 5 u5 + C = 5 tan5 + C. April,

86 65 CHAPTER 5 THE ITEGRAL 57. sec d Let u =. Then du = d or du = d. Hence, sec d = sec ud = tan u + C = tan + C. 58. cos + sin d Let u = + sin. Then du = cos or du = cos d. Hence + sin cos d= u du = u + C = + sin + C. 59. sin cos + d Let u = cos +. Then du = sin or du = sin. Hence sin cos + d = u / du = u/ + C = 6 cos + / + C. 6. cos sin d Let u = sin. Then du = cos dor du = cos d. Hence sin cos d= udu= u + C = 6 sin + C. 6. sec θ tan θsec θ dθ Let u = sec θ. Then du = sec θ tan θdθand sec θ tan θsec θ dθ = udu= u + C = sec θ + C. 6. cos t cossin tdt Let u = sin t. Then du = cos tdtand cos t cossin tdt = cos udu= sin u + C = sinsin t + C. 6. e 7 d Let u = 7. Then du = d or du = d. Hence, e 7 d = e u du = eu + C = e 7 + C e + d 65. Let u = +. Then du = + d or du = + d. Hence, + e + d = e u du = eu + C = e + + C. e d e + Let u = e +. Then du = e d, and e e + d = u du = u + C = e + + C. April,

87 66. sec θe tan θ dθ SECTIO 5.6 Substitution Method 65 Let u = tan θ. Then du = sec θdθ, and sec θe tan θ dθ = e u du = e u + C = e tan θ + C. e t dt 67. e t + e t + Let u = e t. Then du = e t dt, and e t dt e t + e t + = du u + u + = d ln Let u = ln. Then du = d, and d ln = ln d d ln du u + = u + + C = e t + + C. u du = u + C = ln + C. Let u = ln. Then du = d, and ln d = u du = 5 u5 + C = 5 ln 5 + C. Let u = ln. Then du = d, and d ln = du u = ln u +C = ln ln +C. tanln 7. d Let u = cosln. Then du = sinln d or du = sinln d. Hence, tanln sinln du d = d = = ln u +C = ln cosln +C. cosln u 7. cot lnsin d Let u = lnsin. Then and cot lnsin d = du = cos = cot, sin udu= u + C = lnsin + C. 7. Evaluate d + using u = +. Hint: Show that d = u du. Let u = +. Then du = d or d = du= u du. April,

88 65 CHAPTER 5 THE ITEGRAL Hence, d u + = u du = u u du = u + u + C = C. 7. Can The Both Be Right? Hannah uses the substitution u = tan and Akiva uses u = sec to evaluate tan sec d. Show that the obtain different answers, and eplain the apparent contradiction. With the substitution u = tan, Hannah finds du = sec dand tan sec d= udu= u + C = tan + C. On the other hand, with the substitution u = sec, Akiva finds du = sec tan dand tan sec d= sec tan sec d = sec + C Hannah and Akiva have each found a correct antiderivative. To resolve what appears to be a contradiction, recall that an two antiderivatives of a specified function differ b a constant. To show that this is true in their case, note that sec + C tan + C = sec tan + C C = + C C = + C C, a constant Here we used the trigonometric identit tan + = sec. 75. Evaluate sin cos dusing substitution in two different was: first using u = sin and then using u = cos. Reconcile the two different answers. First, let u = sin. Then du = cos dand sin cos d= udu= u + C = sin + C. et, let u = cos. Then du = sin dor du = sin d. Hence, sin cos d= udu= u + C = cos + C. To reconcile these two seemingl different answers, recall that an two antiderivatives of a specified function differ b a constant. To show that this is true here, note that sin + C cos + C = + C C, a constant. Here we used the trigonometric identit sin + cos =. 76. Some Choices Are Better Than Others Evaluate sin cos d twice. First use u = sin to show that sin cos d= u u du and evaluate the integral on the right b a further substitution. Then show that u = cos is a better choice. Consider the integral sin cos d.ifweletu = sin, then cos = u and du = cos d. Hence sin cos d= u u du. ow let w = u. Then dw = uduor dw = udu. Therefore, u u du = w / dw = w/ + C = w/ + C = u / + C = sin / + C = cos + C. April,

89 SECTIO 5.6 Substitution Method 65 A better substitution choice is u = cos. Then du = sin dor du = sin d. Hence sin cos d= u du = u + C = cos + C. 77. What are the new limits of integration if we appl the substitution u = + π to the integral π sin + πd? The new limits of integration are u = + π = π and uπ = π + π = π. 78. Which of the following is the result of appling the substitution u = 9 to the integral 8 9 d? 8 a u du b 8 u du c u du d u du Let u = 9. Then du = d or du = d. Furthermore, when =, u =, and when = 8, u =. Hence 8 9 d = u du. The answer is therefore d. In Eercises 79 9, use the Change-of-Variables Formula to evaluate the definite integral d Let u = +. Then du = d. Hence 5 + d = u du = 5 u = 5 = d Let u = +. Then du = d. Hence d = udu= u/ = = d Let u = +. Then du = dor du = d. Hence + d = u du = u = 6 + = 6 = d Let u = Then du = 5 d or 5 du = d. Hence d = udu= 5 5 u/ = 6 = d Let u = + 9. Then du = dor du = d. Hence + 9 d = 5 5 udu= 9 u/ = = 9. April,

90 65 CHAPTER 5 THE ITEGRAL d Let u = Then du = + 6d and d = u du = 7 8 u 8 = 7 + = d Let u = +. Then du = + d = + d, and d = u 5 du = u6 = 79 = / d Let u = 9. Then du = d. Hence / d = u / d = u / 8 = =. 87. θ tanθ dθ Let u = cos θ. Then du = θ sin θ dθ or du = θ sin θ dθ. Hence, θ tanθ θ sinθ cos dθ = cosθ dθ = du u = cos ln u = [lncos + ln ] = lnsec. π/6 88. sec π d 6 Let u = π. Then du = d and 6 π/6 sec π d = π/6 sec udu= π/6 6 π/6 tan u π/6 = + =. π/ 89. cos sin d Let u = cos. Then du = sin d. Hence π/ cos sin d= u du = u du = u = =. 9. π/ cot π/ csc d Let u = cot. Then du = csc and π/ π/ cot csc d = u du = u =. April,

91 SECTIO 5.6 Substitution Method Evaluate r 5 r dr. Let u = 5 r. Then du = rdr = rdr r 5 u so that Hence, the integral becomes: r 5 r dr = = rdr= 5 u du. 5 5 u5 u du = u / u / du = u/ 5 5 u5/ 8 = Find numbers a and b such that b a π/ u + du= sec θdθ π/ and evaluate. Hint: Use the identit sec θ = tan θ +. Let u = tan θ. Then u + = tan θ + = sec θ and du = sec θdθ. Moreover, because tan π = and tan π =, it follows that a = and b =. Thus, π/ sec θdθ= u + du= π/ u + u = Wind engineers have found that wind speed v in meters/second at a given location follows a Raleigh distribution of the tpe Wv = ve v /6 This means that at a given moment in time, the probabilit that v lies between a and b is equal to the shaded area in Figure. a Show that the probabilit that v [,b] is e b /6. b Calculate the probabilit that v [, 5].. = Wv.5 a b v m/s FIGURE The shaded area is the probabilit that v lies betweena and b. a The probabilit that v [,b] is b ve v /6 dv. Let u = v /6. Then du = v/ dv and b b /6 ve v /6 dv = e u du = e u b /6 = e b /6 +. April,

92 656 CHAPTER 5 THE ITEGRAL b The probabilit that v [, 5] is the probabilit that v [, 5] minus the probabilit that v [, ]. B part a, the probabilit that v [, 5] is e 5/6 e /6 = e /6 e 5/6. π/ 9. Evaluate sin n cos dfor n. Let u = sin. Then du = cos d. Hence π/ sin n cos d= u n du = un+ n + In Eercises 95 96, use substitution to evaluate the integral in terms of f. 95. f f d = n +. Let u = f. Then du = f d. Hence f f d = u du = u + C = f + C. 96. f f d Let u = f. Then du = f d. Hence f f d = u du = u + C = f + C. π/6 / 97. Show that fsin θdθ = f u du. u Let u = sin θ. Then uπ/6 = / and u =, as required. Furthermore, du = cos θdθ, so that dθ = du cos θ. If sin θ = u, then u + cos θ =, so that cos θ = u. Therefore dθ = du/ u. This gives π/6 / fsin θdθ = f u du. u Further Insights and Challenges 98. Use the substitution u = + /n to show that + /n d = n u / u n du Evaluate for n =,. Let u = + /n. Then = u n and d = nu n du. Accordingl, + /n d = n u / u n du. For n =, we have + / d = For n =, we have + / d = u / u du = u / u / du = 5 u5/ u/ + C = 5 + / 5/ + / / + C. u / u du = u 5/ u / + u / du April,

93 = 7 u7/ u 5/ + 5 u/ + C SECTIO 5.6 Substitution Method 657 = / 7/ 5 + / 5/ + + / / + C. π/ 99. Evaluate I = π/ then check that I + J = To evaluate dθ + tan 6, θ dθ. π/. Hint: Use substitution to show that I is equal to J = π/ d I = + tan 6, we substitute t = π/. Then dt = d, = π/ t, t = π/, and tπ/ =. Hence, π/ I = d + tan 6 = π/ Let J = π/ dt + cot 6 t linearit of the integral, π/ d I + J = + tan 6 + d + cot 6 = π/ = + tan / tan 6 dt π/ + tan 6 π/ t = dt + cot 6 t. dθ + cot 6, θ and. We know I = J,soI + J = I. On the other hand, b the definition of I and J and the π/ d π/ = + tan 6 + tan 6 + / tan 6 π/ = + tan 6 + tan6 + tan 6 d π/ + tan 6 π/ = + tan 6 d = d = π/. + tan cot 6 Hence, I + J = I = π/, so I = π/. a. Use substitution to prove that fd = iff is an odd function. a We assume that f is continuous. If f is an odd function, then f = f. Let u =. Then = uand du = d or du = d. Accordingl, a a a fd = fd+ fd = f u du + fd a a a a a = fd f u du =.. Prove that b a d = b/a d for a, b>. Then show that the regions under the hperbola over the intervals [, ], [, ], [, 8],...all have the same area Figure 5. = d d Equal area 8 8 FIGURE 5 The area under = over [n, n+ ] is the same for all n =,,,... April,

94 658 CHAPTER 5 THE ITEGRAL a Let u = a. Then au = and du = a d or adu= d. Hence b b/a a d = a b/a au du = u du. b/a b/a ote that u du = d after the substitution = u. b The area under the hperbola over the interval [, ] is given b the definite integral d. Denote this definite integral b A. Using the result from part a, we find the area under the hperbola over the interval [, ] is / d = d = Similarl, the area under the hperbola over the interval [, 8] is 8 8/ d = d = In general, the area under the hperbola over the interval [ n, n+ ] is d = A. d = A. n+ n+ / n n d = d = d = A.. Show that the two regions in Figure 6 have the same area. Then use the identit cos u = + cos u to compute the second area. = = cos u A FIGURE 6 B u The area of the region in Figure 6A is given b d. Let = sin u. Then d = cos uduand = sin u = cos u. Hence, π/ π/ d = cos u cos udu= cos udu. This last integral represents the area of the region in Figure 6B. The two regions in Figure 6 therefore have the same area. Let s now focus on the definite integral π/ cos udu. Using the trigonometric identit cos u = + cos u,we have π/ cos udu= π/ + cos udu= u + π/ sin u = π = π.. Area of an Ellipse Prove the formula A = πab for the area of the ellipse with equation Figure 7 a + b = Hint: Use a change of variables to show that A is equal to ab times the area of the unit circle. b a a b FIGURE 7 Graph of a + b =. April,

95 SECTIO 5.7 Further Transcendental Functions 659 Consider the ellipse with equation a + b = ; here a,b >. The area between the part of the ellipse in the upper half-plane, = f= b a, and the -ais is a a fd. B smmetr, the part of the elliptical region in the lower half-plane has the same area. Accordingl, the area enclosed b the ellipse is a a fd = b a a a a d = b /a d a ow, let u = /a. Then = au and adu= d. Accordingl, a b a d = ab a π u du = ab = πab Here we recognized that u du represents the area of the upper unit semicircular disk, which b Eercise is π = π. 5.7 Further Transcendental Functions Preliminar Questions b d. Find b such that is equal to a ln b For b>, b d b = ln = ln b ln = ln b. a For the value of the definite integral to equal ln, we must have b =. b For the value of the definite integral to equal, we must have b = e. b d. Find b such that + = π. In general, b d + = b tan = tan b tan = tan b. For the value of the definite integral to equal π, we must have tan b = π or b = tan π =.. Which integral should be evaluated using substitution? 9 d a + b Use the substitution u = on the integral in b. d + 9. Which relation between and u ields 6 + = + u? To transform 6 + into + u, make the substitution = u. Eercises In Eercises, evaluate the definite integral. 9. d. d 9 9 d = ln = ln 9 ln = ln 9. d = ln = ln ln = ln 5. April,

96 66 CHAPTER 5 THE ITEGRAL e. t dt e e. e t dt 5. t e e t dt t + e dt = ln t = ln e ln =. e dt = ln t = ln e ln e =ln e = ln/e =. e e Let u = t +. Then du = dt and dt t + = du u = ln u = ln ln = ln. e dt 6. e t ln t Let u = ln t. Then du = /tdt and e e t ln t dt = du u = ln u = ln ln = ln. tan 8 d 7. tan + tan 8 d tan + = tan 8 tan = tan tan 8 tan tan = 8 = 7. tan 7 d 8. + Let u = +. Then du = dand 7 d + = 5 du 5 u = 5 ln u = 5 ln 5 ln 5 = ln. / d 9. / d = / sin = sin sin = π 6. / d. / d = / sec. Use the substitution u = / to prove = sec sec = 5π 6 π = π 6. d 9 + = tan + C Let u = /. Then, = u, d = du,9+ = 9 + u, and d 9 + = du 9 + u = du + u = tan u + C = tan + C.. Use the substitution u = to evaluate d +. Let u =. Then, = u/, d = du, + = u +, and d + = du u + = tan u + C = tan + C. April,

97 SECTIO 5.7 Further Transcendental Functions 66 In Eercises, calculate the integral. d. + Let = u. Then d = du and d + = du u + = tan u = tan tan = π. dt. t + 9 Let t = /u. Then dt = /du,t + 9 = 9t + 9 = 9t +, and dt t + 9 = 8/ du 6 u + = 8/ 6 tan u = 8 6 tan. dt 5. 6t Let u = t. Then du = dt, and dt = 6t du u = sin u + C = sin t + C. /5 d 6. /5 5 Let = u/5. Then and 7. dt 5 t /5 /5 d = 5 du, 5 = u, d = / 5 5 / u du = / 5 sin u = 5 / sin sin = π 5. Let t = 5/u. Then dt = 5/ du and dt 5/ du = = du = sin u + C = sin 5 t 5 t u 5 t + C. / d 8. / 6 Let = u/. Then d = du/, 6 = u and / d / 6 = du u u = sec u = sec sec = π. d 9. Let u =. Then du = d and d = du u u = sec u + C = sec + C. April,

98 66 CHAPTER 5 THE ITEGRAL d. + Let u =. Then du = dand d + = du u + = tan u + C = tan + C. d. Let u =. Then du = d, and d = du u u = sec u + C = sec + C.. / + d Observe that + d = d + d. In the first integral on the right, we let u =, du = d. Thus + d = du u / + d = + sin + C. Finall, + d = + sin / = + / + π 6. e d. ln + e. Let u = e. Then du = e d, and e d ln + e = du / + u = tan u = π / tan /. lncos d cos Let u = ln cos. Then du = 5. tan d + lncos d cos = cos, and udu= u + C = ln cos + C. Let u = tan. Then du = d +, and tan d + = udu= u + C = tan d 6. tan + Let u = tan. Then du = d +, and d π/ tan + = π/ du = ln u π/ u = ln π π/ ln π = ln. + C. April,

99 7. d d = ln = = ln ln. 8. d Let u =. Then du = d and d = u du = u ln SECTIO 5.7 Further Transcendental Functions 66 = ln + = ln. log 9. d log d = log ln = = ln ln = ln.. t5 t dt Let u = t. Then du = t dtand t5 t dt = 5 u du = 5u ln5 = 5 ln5 ln5 = ln sin9 d. Let u = 9. Then du = 9 ln 9 d and 9 sin9 d = sin udu= cos u + C = ln 9 ln 9 ln 9 cos9 + C. d 5 First, rewrite d 5 = d 5 5 = 5 d 5. ow, let u = 5. Then du = 5 ln 5 d and d 5 = du ln 5 = u ln 5 sin u + C = ln 5 sin 5 + C. In Eercises 7, evaluate the integral using the methods covered in the tet so far.. e d Use the substitution u =,du= d. Then e d = e u du = eu + C = e + C.. 5. d + 5 Let u = + 5. Then du = d and d + 9 d + 5 = du u = ln u +C = ln + 5 +C. Let u = + 9. Then du = 8dand + 9 d = u / du = 8 u/ + C = C. April,

100 66 CHAPTER 5 THE ITEGRAL 6. d d = 7. 7 d + d = ln + C. 8. e 9 t dt Let u =. Then du = d and 7 d = 7 u du = 7u ln 7 + C = 7 ln 7 + C. Let u = 9 t. Then du = dt and e 9 t dt = e u du = eu + C = e9 t + C. 9. sec θ tan 7 θdθ Let u = tan θ. Then du = sec θdθand sec θ tan 7 θdθ= u 7 du = 8 u8 + C = 8 tan8 θ + C... cosln tdt t tdt 7 t Let u = ln t. Then du = dt/t and cosln tdt = t cos udu= sin u + C = sinln t + C. Let u = 7 t. Then du = t dtand tdt = u / du = u / + C = 7 t 7 t + C.. e d First, note that Thus, = e ln so e = e +ln. e d = e +ln d = + d. + Write + d = + In the first integral, let u = +. Then du = dand d + = + ln e+ln + C. d + + d +. du u ln u +C = ln + + C. For the second integral, let = u. Then d = du and d + = du u + = tan u + C = tan / + C. April,

101 SECTIO 5.7 Further Transcendental Functions 665 Combining these two results ields. tan + d + d + = ln + + tan / + C. and 5. First we rewrite tan + d as sin+ cos+ d. Let u = cos +. Then du = sin + d, d 6 sin + cos + d = du u = ln cos + +C. Let u =. Then du = d and d = du 6 = u sin u + C = sin + C. 6. e t e t + dt 7. e d Use the substitution u = e t +,du= e t dt. Then e t udu= e t + dt = u/ + C = et + / + C. First, observe that e d = e d d= e d. In the remaining integral, use the substitution u =,du = d. Then e d = e u du = e u + C = e + C. Finall, e d = e + C e d First, observe that 7 e d = 7 d e d = 7 e d. In the remaining integral, use the substitution u =,du = d. Then e d = e u du = eu + C = e + C. Finall, 7 e d = 7 e + C. e e 9. e d e e e d = e e d = e e + C. April,

102 666 CHAPTER 5 THE ITEGRAL d 5. 5 Let u = 5. Then du = 5 d and d 5 = d du u u = sec u + C = sec 5 + C. Write + 5d = d + 5 d. In the first integral, let u =. Then du = dand d = u / du = u / + C = + C. In the second integral, let = u. Then d = du and 5 d = 5 du = 5 u sin u + C = 5 sin / + C. Combining these two results ields + 5d = + 5 sin / + C. 5. t + t + dt Let u = t +. Then du = dt and t + t + dt = u / du = 5 u5/ + C = 5 t + 5/ + C. 5. e cose d Use the substitution u = e,du= e d. Then e cose d = cos udu= sin u + C = sine + C. 5. e e + d Use the substitution u = e +,du= e d. Then e du e + d = = u + C = e + + C. u 55. d 9 6 First rewrite d = 9 6 d. ow, let u =. Then du = d and d = 9 6 du = u sin u + C = sin + C. April,

103 56. d ln8 Let u = ln8. Then du = 8 8 d = 57. e e + d d ln8 = SECTIO 5.7 Further Transcendental Functions 667 d, and du u = ln u +C = ln ln8 +C. Use the substitution u = e,du= e d. Then e e + d = u + du = u 6 + u + u + du 58. d ln 5 = 7 u7 + 5 u5 + u + u + C = 7 e e 5 + e + e + C = e7 7 + e5 + e + e + C. 5 Let u = ln. Then du = d/ and d ln 5 = u 5 du = u + C = ln + C. d Let u = +. Then du = d, and d + = 6. d 9 + du u = ln + +C. Let u = 9 +. Then du = + 6d = d, and d 9 + = du u = ln9 + + C. 6. cot d 6. We rewrite cot das cos sin d. Let u = sin. Then du = cos d, and cos sin d = du = ln sin +C. u cos sin + d Let u = sin +. Then du = cos d, and cos sin + d = du u = ln sin + + C, where we have used the fact that sin + to drop the absolute value. ln d Let u = ln + 5. Then du = /d, and ln + 5 d = udu= 8 u + C = 8 ln C. April,

104 668 CHAPTER 5 THE ITEGRAL 6. sec θ tan θ5 sec θ dθ Let u = sec θ. Then du = sec θ tan θdθand sec θ tan θ5 sec θ dθ = 5 u du = 5u ln 5 + C = 5sec θ ln 5 + C. 65. d Let u =. Then du = d, and d = u du = u ln + C = ln + C. lnln 66. ln d Let u = lnln. Then du = ln lnln ln d = 67. cot lnsin d d and udu= u + C = lnln + C. Let u = lnsin. Then and 68. tdt t cot lnsin d = du = cos d= cot d, sin udu= u + C = lnsin + C. Let u = t. Then du = t dtand tdt = du t = u sin u + C = sin t + C. 69. t t dt Let u = t. Then t = u +, du = dt and t t dt = u + udu = u + 6u + 9 udu= u 5/ + 6u / + 9u / du = 7 u7/ + 5 u5/ + 6u / + C = 7 t 7/ + 5 t 5/ + 6t / + C. 7. cos 5 sin d Let u = sin. Then du = cos dand cos 5 sin d = 5 u du = 5u sin ln5 + C = 5 ln5 + C. April,

105 SECTIO 5.7 Further Transcendental Functions Use Figure to prove t dt = + sin FIGURE The definite integral t dt represents the area of the region under the upper half of the unit circle from to. The region consists of a sector of the circle and a right triangle. The sector has a central angle of π θ, where cos θ =. Hence, the sector has an area of π cos = sin. The right triangle has a base of length, a height of, and hence an area of. Thus, t dt = + sin. 7. Use the substitution u = tan to evaluate Hint: Show that If u = tan, then du = sec dand d + sin. d + sin = du + u Thus du + u = sec d + tan = d cos + sin = d + sin = du + u = d cos + sin + sin = d + sin. du + u = tan u + C = tan tan + C. 7. Prove: sin tdt= t + t sin t. Let Gt = t + t sin t. Then G t = d t dt + d t sin t = t dt + t d t dt sin t + sin t = t + t + t t sin t = sin t. This proves the formula 7. a Verif for r : sin tdt= t + t sin t. T te rt dt = ert rt + r 6 Hint: For fied r, let FTbe the value of the integral on the left. B FTC II, F t = te rt and F =. Show that the same is true of the function on the right. b Use L Hôpital s Rule to show that for fied T, the limit as r of the right-hand side of Eq. 6 is equal to the value of the integral for r =. April,

106 67 CHAPTER 5 THE ITEGRAL a Let Then and as required. b Using L Hôpital s Rule, e rt rt + lim r r T T If r = then, te rt dt = tdt= t ft= ert rt + r r. f t = r e rt r + rt re rt = te rt f = r + r =, Te rt + rt T e rt = lim r r T = T. Further Insights and Challenges 75. Recall that if ft gt for t, then for all, rt e rt = lim r r T e rt = lim r = T. ftdt gt dt 7 The inequalit e t holds for t because e>. Use Eq. 7 to prove that e + for. Then prove, b successive integration, the following inequalities for : e + +, e Integrating both sides of the inequalit e t ields e t dt = e or e +. Integrating both sides of this new inequalit then gives e t dt = e + / or e + + /. Finall, integrating both sides again gives e t dt = e + / + /6 or e + + / + /6 as requested. 76. Generalize Eercise 75; that is, use induction if ou are familiar with this method of proof to prove that for all n, e n! n For n =, e + b Eercise 75. Assume the statement is true for n = k. We need to prove the statement is true for n = k +. B the Induction Hpothesis, e + + / + + k /k!. Integrating both sides of this inequalit ields e t dt = e + / + + k+ /k +! or as required. e + + / + + k+ /k +! April,

107 SECTIO 5.7 Further Transcendental Functions Use Eercise 75 to show that e / /6 and conclude that lim e / =. Then use Eercise 76 to prove more generall that lim e / n = for all n. B Eercise 75, e Thus e Since lim /6 =, lim e / =. More generall, b Eercise 76, e n+ n +!. Thus e n n + + n +! n +!. Since lim n+! =, lim e n =. Eercises 78 8 develop an elegant approach to the eponential and logarithm functions. Define a function G for >: G = t dt 78. Defining ln as an Integral This eercise proceeds as if we didn t know that G = ln and shows directl that G has all the basic properties of the logarithm. Prove the following statements. a ab a t dt = b t dt for all a,b >. Hint: Use the substitution u = t/a. b Gab = Ga + Gb. Hint: Break up the integral from to ab into two integrals and use a. c G = and Ga = Ga for a>. d Ga n = nga for all a>and integers n. e Ga /n = Ga for all a>and integers n. n f Ga r = rga for all a>and rational numbers r. g G is increasing. Hint: Use FTC II. h There eists a number a such that Ga >. Hint: Show that G > and take a = m for m>/g. i lim G = and lim G = + j There eists a unique number E such that GE =. k GE r = r for ever rational number r. a Let u = t/a. Then du = dt/a, ua =, uab = b and b Using part a, ab a ab t dt = a b a at dt = b u du = t dt. ab a Gab = t dt = ab t dt + a a t dt = b t dt + dt = Ga + Gb. t c First, et, Ga = G /a = a a = G = dt =. t t dt = a t dt dt = Ga. t b part a with b = a April,

108 67 CHAPTER 5 THE ITEGRAL d Using part a, a n Ga n = a = a t dt = a t dt + a t dt + a dt + + t e Ga = Ga /n n = nga /n. Thus, Ga /n = n Ga. f Let r = m/n where m and n are integers. Then t a Ga r = Ga m/n = Ga m /n = n Gam b part e = m Ga b part d n = rga. a n dt + + a n t dt dt = nga. t g B the Fundamental Theorem of Calculus, G is continuous on, and G = is increasing and one-to-one for >. h First note that > for >. Thus, G G = t dt > > because t > for t,. ow, let a = m for m an integer greater than /G. Then Ga = G m = mg > G =. G i First, let a be the value from part h for which Ga > note that a itself is greater than. ow, lim G = m lim Gam = Ga lim m m =. For the other limit, let t = / and note lim G = lim + t G = lim Gt =. t t j B part c, G = and b part h there eists an a such that Ga >. the Intermediate Value Theorem then guarantees there eists a number E such that <E<aand GE =. We know that E is unique because G is one-to-one. k Using part f and then part j, GE r = rge = r = r. 79. Defining e Use Eercise 78 to prove the following statements. a G has an inverse with domain R and range { : >}. Denote the inverse b F. b F + = FF for all,. Hint: It suffices to show that GF F = GF +. c Fr = E r for all numbers. In particular, F =. d F = F. Hint: Use the formula for the derivative of an inverse function. This shows that E = e and Fis the function e as defined in the tet. a The domain of G is >and, b part i of the previous eercise, the range of G is R. ow, G = > for all >. Thus, G is increasing on its domain, which implies that G has an inverse. The domain of the inverse is R and the range is { : >}. Let Fdenote the inverse of G. b Let and be real numbers and suppose that = Gw and = Gz for some positive real numbers w and z. Then, using part b of the previous eercise F + = F Gw + Gz = F Gwz = wz = F+ F. April,

109 SECTIO 5.7 Further Transcendental Functions 67 c Let r be an real number. B part k of the previous eercise, GE r = r. B definition of an inverse function, it then follows that Fr = E r. d B the formula for the derivative of an inverse function F = G F = /F = F. 8. Defining b Let b>and let f = F Gb with F as in Eercise 79. Use Eercise 78 f to prove that fr= b r for ever rational number r. This gives us a wa of defining b for irrational, namel b = f. With this definition, b is a differentiable function of because F is differentiable. B Eercise 78 f, fr= F rgb = F Gb r = b r, for ever rational number r. 8. The formula n d = n+ + C is valid for n. Show that the eceptional case n = is a limit of the n + general case b appling L Hôpital s Rule to the limit on the left. lim t n dt = t dt for fied > n ote that the integral on the left is equal to n+ n +. lim t n t n+ dt = lim n n n + = lim n n+ n + n+ n + n+ = lim = lim n n + n n+ ln = ln = t dt ote that when using L Hôpital s Rule in the second line, we need to differentiate with respect to n. 8. The integral on the left in Eercise 8 is equal to f n = n+. Investigate the limit graphicall b n + plotting f n for n =,.,.6, and.9 together with ln on a single plot. n = n =. n =.6 n =.9 = ln 5 8. a Eplain wh the shaded region in Figure 5 has area ln a e d. b Prove the formula a ln d= a ln a ln a e d. c Conclude that a ln d= a ln a a +. d Use the result of a to find an antiderivative of ln. = ln ln a a FIGURE 5 April,

110 67 CHAPTER 5 THE ITEGRAL a Interpreting the graph with as the independent variable, we see that the function is = e. Integrating in then gives the area of the shaded region as ln a e d b We can obtain the area under the graph of = ln from = to = a b computing the area of the rectangle etending from = to = a horizontall and from = to = ln a verticall and then subtracting the area of the shaded region. This ields c B direct calculation Thus, a ln a ln d= a ln a e d. ln a e d = e ln a = a. a ln d= a ln a a = a ln a a +. d Based on these results it appears that ln d= ln + C. 5.8 Eponential Growth and Deca Preliminar Questions. Two quantities increase eponentiall with growth constants k =. and k =., respectivel. Which quantit doubles more rapidl? Doubling time is inversel proportional to the growth constant. Consequentl, the quantit with k =. doubles more rapidl.. A cell population grows eponentiall beginning with one cell. Which takes longer: increasing from one to two cells or increasing from 5 million to million cells? It takes longer for the population to increase from one cell to two cells, because this requires doubling the population. Increasing from 5 million to million is less than doubling the population.. Referring to his popular book A Brief Histor of Time, the renowned phsicist Stephen Hawking said, Someone told me that each equation I included in the book would halve its sales. Find a differential equation satisfied b the function Sn, the number of copies sold if the book has n equations. ields Let S denote the sales with no equations in the book. Translating Hawking s observation into an equation Sn = S n. Differentiating with respect to n then ields ds dn = S d dn n = ln S n = ln Sn.. The PV of dollars received at time T is choose the correct answer: a The value at time T of dollars invested toda b The amount ou would have to invest toda in order to receive dollars at time T The correct response is b: the PV of dollars received at time T is the amount ou would have to invest toda in order to receive dollars at time T. 5. In one ear, ou will be paid $. Will the PV increase or decrease if the interest rate goes up? If the interest rate goes up, the present value of $ a ear from now will decrease. April,

111 SECTIO 5.8 Eponential Growth and Deca 675 Eercises. A certain population P of bacteria obes the eponential growth law Pt = e.t t in hours. a How man bacteria are present initiall? b At what time will there be, bacteria? a P = e = bacteria initiall. b We solve e.t =, for t. Thus, e.t = 5or t = ln 5. hours... A quantit P obes the eponential growth law Pt = e 5t t in ears. a At what time t is P =? b What is the doubling time for P? a e 5t = when t = 5 ln.6 ears. b The doubling time is 5 ln. ears.. Write ft= 57 t in the form ft= P e kt for some P and k. Because 7 = e ln 7, it follows that ft= 57 t = 5e ln 7 t = 5e t ln 7. Thus, P = 5 and k = ln 7.. Write ft= 9e.t in the form ft= P b t for some P and b. Observe that ft= 9e.t = 9 e. t, so P = 9 and b = e A certain RA molecule replicates ever minutes. Find the differential equation for the number t of molecules present at time t in minutes. How man molecules will be present after one hour if there is one molecule at t =? The doubling time is ln ln so k = k doubling time. Thus, the differential equation is t = kt = ln t. With one molecule initiall, t = e ln /t = t/. Thus, after one hour, there are 6 = 6/ =,8,576 molecules present. 6. A quantit P obes the eponential growth law Pt = Ce kt t in ears. Find the formula for Pt, assuming that the doubling time is 7 ears and P =. The doubling time is 7 ears, so 7 = ln /k,ork = ln /7 =.99 ears. With P =, it follows that Pt = e.99t. 7. Find all s to the differential equation = 5. Which satisfies the initial condition =.? = 5, sot = Ce 5t for some constant C. The initial condition =. determines C =.. Therefore, t =.e 5t. 8. Find the to = satisfing =. =, sot = Ce t for some constant C. The initial condition = determines C =. Therefore, t = e t. 9. Find the to = satisfing =. =, sot = Ce t for some constant C. The initial condition = determines C = e 6. Therefore, t = e 6 et = e t. April,

112 676 CHAPTER 5 THE ITEGRAL. Find the function = ftthat satisfies the differential equation =.7 and the initial condition =. Given that =.7 and =, then ft= e.7t.. The deca constant of cobalt-6 is. ear. Find its half-life. Half-life = ln 5. ears... The half-life radium-6 is 6 ears. Find its deca constant. Half-life = ln ln so k = k half-life = ln 6 =.7 ears.. One of the world s smallest flowering plants, Wolffia globosa Figure, has a doubling time of approimatel hours. Find the growth constant k and determine the initial population if the population grew to after 8 hours. FIGURE The tin plants are Wolffia, with plant bodies smaller than the head of a pin. B the formula for the doubling time, = ln k. Therefore, k = ln. hours. The plant population after t hours is Pt = P e.t.ifp8 =, then P e.8 = P = e.8. A -kg quantit of a radioactive isotope decas to kg after 7 ears. Find the deca constant of the isotope. Pt = e kt. Thus P7 = = e 7k,sok = ln/.7 ears The population of a cit is Pt = e.6t in millions, where t is measured in ears. Calculate the time it takes for the population to double, to triple, and to increase seven-fold. Since k =.6, the doubling time is ln.55 ears. k The tripling time is calculated in the same wa as the doubling time. Solve for in the equation Pt + = Pt e.6t+ = e.6t e.6t e.6 = e.6t e.6 =.6 = ln, or = ln /.6 8. ears. Working in a similar fashion, we find that the time required for the population to increase seven-fold is ln 7 k = ln 7. ears What is the differential equation satisfied b Pt, the number of infected computer hosts in Eample? Over which time interval would Ptincrease one hundred-fold? Because the rate constant is k =.85 s, the differential equation for Ptis dp dt =.85P. April,

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