Math b Review Sheet for Eam SOLUTIONS The first Math b midterm will be Tuesday, February 8th, 7 9 p.m. Location: Schwartz Auditorium Room ) The eam will cover: Section 3.6: Inverse Trig Appendi F: Sigma Notation Section 5.: Areas and Distances Section 5.: The Definite Integral Section 5.4: Fundamental Theorem of Calculus Part I) No calculators will be allowed on the eam. Our eams are closed book: no notes, books, calculators, cell phones or internetconnected devices will be allowed. On the eam, you will be asked to show all your work. We are grading your ability to clearly communicate your understanding of how to solve each problem much more than your ability to get the correct answer. In fact, a correct answer with little or no work might not receive any points at all. Showing your work clearly gives us justification for assigning partial credit. On the eam, you will be given the following three summation formulae so you do *not* need to memorize these): i = nn + ) i = nn + )n + ) 6 [ nn + ) i 3 = ] The solutions to this review material will be available on LATTE by Tue, Feb 4. OVER for Practice Eam SOLUTIONS
Practice Eam SOLUTIONS Problems - ) NOTE: You may use the following three summation formulae whenever needed. These formulae will also be given to you on the eam. [ ] nn + ) nn + )n + ) nn + ) ) i =, ) i =, 3) i 3 = 6. Below is the graph of f). 3 3 4 5 6 7 3 a) Find 7 f)d. Note that the part of the graph between = 5 and = 7 is a semi-circle.) Solution: 7 f)d = f)d + f)d + f)d + 5 = )) )) )) π) )) + = π 4 3 f)d + 7 5 f)d b) Using your answer from part a, find 7 f) + 5 d. Solution: 7 f) + 5 d = 5 7 f) d + 7 = 5 π 4) + 7 ) 5 5 d = π +
. Compute the following: a) sin ) 3 Solution: Since sin ) π 3 = 3, we have sin ) 3 = π. 3 b) arctan tan )) 7π 4 Solution: arctan tan )) 7π 4 = arctan ) = π 4 c) cos arcsin )) 5 3 Solution: Let θ = arcsin ) 5 3. Then sin θ = 5. 3 5 3 a θ So cos arcsin )) 5 3 = cos θ = 3 a + 5 = 3 a = 69 5 a = 44 a = 3. Find the derivatives of the following functions: a) f) = lnarctan)) Solution: f ) = arctan) ) ) ) + 4 b) h) = sint ) dt
Solution: h) = = = sint ) dt sint ) dt + sint ) dt + sint ) dt sint ) dt Since ft) = sint ) is continuous on, ), we can use part I of the Fundamental Theorem of Calculus FTC I). By FTCI, we know d d sint ) dt ) = sin ). For the second integral, we use the chain rule together with FTC I where our inside function is g) =, so g ) = ) and so d ) sint ) dt = sin 4 ). d So h ) = sin ) + sin 4 ) 4. Evaluate the following: 4 k= 4k 3). Your answer should be a number.) Solution : Let i = k + 3, so k = i 3. Then 4 7 4k 3) = 4i 3) 3) k= 7 = 4i 5 7 7 = 4 i 5 ) 77 + ) = 4 75) = 47)4) 75) = 756 5) = 74) = 7
Solution : Here s another way you can do the problem: 4 4 4k 3) = 4 ) 3) + 4 ) 3) + 4) 3) + 4k 3) k= k= 4 4 = 7 3 + 4 k 3 k= k= = + 4 45) 34) = + 45 3) = + 447) = 7 Note: there are some other variations as well. 5. Consider the definite integral + d. a) Write out the sigma notation with i, etc.) for the left-hand Riemann sum L 3 for this integral. Then find and i and finally plug them in to compute L 3. Solution: L 3 = f i ) = f ) + f ) + f ) i= = a b n = 3 3 = i = a + i = + i) = i L 3 = + )) + + )) + + )) = 4 + + 4 = 9 4 b) On the aes below, carefully sketch the graph of f) = +, shading in the area represented by the above integral. Then draw in the rectangles that represent
the area you computed in 5a. y 3 c) Is your estimate in 5a an overestimate or an underestimate of + d, or is there not enough information to determine one way or the other? Briefly eplain your answer. Our estimate is an overestimate since the function is decreasing and we are using left-hand end-points for the height of the rectangles so the y-values of the function over the remainder of each subinterval are less than the y-value we re using for the height of the rectangle for each subinterval). 6. We will compute the volume of a sphere using calculus! z z The volume of this solid sphere can be found by taking the following integral: π z ) dz Compute this integral using the limit definition of the integral to find the volume of the sphere of radius one! Solution: z = b a n = ) n = n z i = a + i z = + i n
) π z dz = π = π lim n z dz [ = π lim n n = π lim n n = π lim n n [ 8 = π lim n n [ = π lim 4 n = π 4 8 ) 3 + i n ) ] n ) 4i n + 4i n ) 4i n 4i n ) 8 i 8 i n 3 ) nn + ) 8 )] nn + )n + ) n 3 6 ) n + 4 )] n + 3n + n 3 n = 4π 3 7. Write the following sum in sigma notation with inde starting at. 3 + 5 7 + 9 + 3 5 Answer: 7 ) n n + ) n= 8. Cameron is training for a marathon. Suppose the function vt) gives Cameron s velocity at time t, where velocity is given in miles per hour and the time is given in t hours since Cameron left home. Cameron s training route is down a straight road.) a) Suppose we know the following information about vt). t 3 vt) 3 6 5 5 4 Use a right-hand Riemann sum with four intervals to estimate vt) dt.
Solution: t = b a n = = 4 4 vt) dt R 4 = vt i ) t = 6 ) = mile + 5 t i = a + i t = + i = i ) 5 ) 4 ) b) What does vt) dt represent in terms of the scenario described above? Describe in words.) Answer: This integral represents how far Cameron is from home after running for two hours. c) What does vt) dt represent in terms of the scenario described above? Describe in words; please do not compute it.) Answer: This integral represents the total distance Cameron traveled after running for two hours. 9. Find the -values of the local etrema for the function f) = t + 3t + dt. Identify each as a local maimum or a local minimum. Solution: Polynomials are continuous, so by FTCI, f ) = + 3 +. f ) = + 3 + = + ) + ) = So the critical numbers are = and =. f ) = + 3 f ) = <, so = is a local maimum f ) = >, so = is a local minimum
. Below are three true or false questions. For each, determine whether the statement is true or false. If a statement is true, briefly eplain why. If a statement is false, give an eample a function or a formula or a graph) that shows that it s false. Let f) be a continuous function on the real numbers, and let g) = ft)dt a) True or false: If f is increasing at = then g) must be positive. FALSE: 3 f This function f is increasing at =, but g) is the negative of the area between the -ais and f, and so g) is negative. b) True or false: If f ) > then g is increasing at = FALSE: The same eample for part a will work here. Since f is increasing at =, we have f ) >, but as increases from =, the function g is decreasing since f is negative. c) True or false: If f) for all, then g) for all. FALSE: Take f) =. Then f is negative on, ). But so g 3) >. g 3) = 3 d = d = ) 3)) = 3, 3 OVER for Practice Eam SOLUTIONS When we say a statement is true we mean it must always be true. When we say a statement is false we mean it could possibly be false.
Practice Eam SOLUTIONS Problems - 4) NOTE: You may use the following three summation formulae whenever needed. These formulae will also be given to you on the eam. [ ] nn + ) nn + )n + ) nn + ) ) i =, ) i =, 3) i 3 = 6. Evaluate the following: a) sinarctan 3 )) = sin π 6 ) = b) arcsinsin π) = sin ) =. Find the domain of f) = lnarcsin)). The domain of arcsin) is [, ]. Since f) = lnarcsin)), we must have arcsin) >, so >. Thus the domain of f) is, ]. 3. Compute the following limit lim h tan 3 + h) tan 3). h Let f) = tan ). Then lim h tan 3 + h) tan 3) h = f 3) = + 3 =. 4. Find the equation of the line tangent to the graph of f) = arctane ) at =. f) = arctane ) f ) = + e ) e. So the slope of the tangent line is f ) = =. The point of tangency is, f)) =, π ). So the equation of the 4 tangent line is y π = ), which can also be written as y = + π. 4 4 5. One or more of these functions is an antiderivative of f) = +. Which ones)? F ) is an antiderivative of f) if F ) = f). So differentiate each of the four functions and see which has derivative f) = +. a) b) d d d d ) ln + ) = ) + ln + = = + = +. +.
c) d) d ) tan = tan + d d d ) tan ) = + = tan + + ) = + 4. +. So F ) = ln + ) is an antiderivative of f), and the other functions are not. 6. Write the following sum using sigma notation. There is more than one correct answer.) 3 + 3 4 4 5 + + 49 5. 49 One correct answer is ) i+ ii + ). 48 Another correct answer is ) i i + )i + ). i= i= 5 Yet another is ) i i )i). 7. Let f) = 4. Estimate the area under the graph of f) over [, ] by finding a) the Riemann sum R 4. Is this an overestimate or underestimate? b) the Riemann sum L 4. Is this an overestimate or underestimate? c) Find the eact area by computing lim n R n. f) = 4. The four subintervals are [, ], [, ], [, 3 ], [ 3, ]. Note that =. a) R 4 = 4 f i ) = f/) + f) + f3/) + f) = 5 4 + 3 + 7 4 + = 7 4. This is an underestimate.
b) L 4 = 4 f i ) = f) + f/) + f) + f3/) = 4 + 5 4 + 3 + 7 4 = 5 4. This is an overestimate. c) Since a = and b =, = =. n n i = + i = i. So we get the n Riemann sum n 4 i. Simplifying gives 4 i ) n ) = n n n ) ) n 4 4i n ) = n 4 n 4i n = 8 n = 8 n 8 nn+)n+) = 8 8n3 +n +4n. n n 3 6 3n 3 ) Then the eact area is lim 8 8n3 +n +4n = 8 8 = 6. n 3n 3 3 3 8 n 3 i 8. Let f) on the interval [, 5]. Suppose that we estimate the area under the graph of y = f) using R 5 and get R 5 = 8. Now suppose that you want to estimate the the area under the graph of the function g) = 3 + f) on the interval [, 5]. What is R 5 for g)? Solution: For f), R 5 = f) + f) + f3) + f4) + f5) = 8. Since =, we can write this as follows: f) + f) + f3) + f4) + f5) = 8. For g), R 5 = g) + g) + g3) + g4) + g5). Since g) = 3 + f) and =, we can rewrite this sum as [ ] [ ] [ ] [ ] [ ] 3 + f) + 3 + f) + 3 + f3) + 3 + f4) + 3 + f5) [ ] = 3 + 3 + 3 + 3 + 3 + [ f) + f) + f3) + f4) + f5) ] [ ] = 3 5 + f) + f) + f3) + f4) + f5). We know that the sum in brackets equals 8, so the entire epression equals 5 + 8) = 3. So R 5 for g) equal 3. 9. Epress the following limit as a definite integral. Do not evaluate the integral. Note that there are many correct answers. lim n sin π + πi n) π n.
We have to choose a value of a. In this problem it looks like = π πi. So π + = n n π + i is good choice for i, which would make a = π. Note: there are other choices for a, but this is the most obvious.) Since = b a and a = π, we get that b π = π n ), so b = π 3π. Finally, the function in the integrand is sin + πi so f n i ) = sin i ), so f) = sin. Hence the integral is π/ π sin d.. Is the following statement true or false? Eplain your answer. False. The integral 4) d represents the area of the region between the graph of f) = 4 and the -ais over the interval [, 3]. 4) d = 3, which is clearly not the area of the region since area must be positive. Instead, 4) d represents the signed or net area between the graph of f) = 4 and the -ais over the interval [, 3]. In other words, ) 4) d = area between the graph of f) and the -ais over, ) ) + area between the graph of f) and the -ais over, 3).. Let f and g be continuous functions. Suppose we know that f) d = ; f) d = 3; g) d = ; g) d = and g) d = 5. Some of the integrals listed below can be evaluated using this information along with properties of integrals and others can t. Find the ones that can, and evaluate them. a) b) c) d) e) f) + g)) d = f) d = f) d + f) d + g) d = + = 3. f) d = + 3 =. f)g) d can t be computed from given information. g) d = 5f)+) d = 5 g) d f) d+ g) d = 5 = 9. d = 5 3)+ ) = 5 3)+ = 4.
f) g). Let F ) = d can t be computed from given information. g) f ) d = y = if < ) f) d since the graph of y = f ) is simply the graph y = if < < ) of y = f) shifted to the right by one unit. So y = if < < 3) if =!! ) ) if 3 < < 5) if < ft) dt, where f) =. =! + 5 if > 3) if < 3 5 if 3 y 4 y a) Sketch the graph of f). f ) d = 3. b) Use the graph to evaluate F ), F ), F 4) and F 6). Remember that F ) is the net i.e., signed) area under the graph of ft) over [, ]. Computing net areas under the graph of f), we get F ) = F ) = F 4) = F 6) = 4 6 ft) dt = ) ) = ft) dt = + )) + )) = ft) dt = + )) + ) + ) = 4.5 ft) dt = 4.5 + )) )) = 4.5 c) Find F ) and F ). F ) = f) =. F ) = f ) =. d) At what points) in the interval [, 6] does F ) =? F ) = when = and when =.5. e) At what point in [, 6] does F ) achieve its absolute i.e., global) maimum value? F ) achieves its global ma in [, 6] at = 5.
f) At what point in [, 6] does F ) achieve its absolute i.e., global) minimum value? F ) achieves its global min in [, 6] at =. g) Is F ) concave up or down at = 4? Why? F ) is concave down at = 4 since F ) is decreasing at = 4. 3. We cut a circular disk of radius r into n circular sectors, as shown in the figure, by marking the angles θ i at which we make the cuts θ = θ n can be considered to be the angle ). A circular sector between two angles θ i and θ i+ has area r θ where θ = θ i+ θ i. We let A n = n i= r θ. Then the area of the disk, A, is given by: a) A n, independent of how many sectors we cut the disk into. b) lim n A n c) π r dθ d) all of the above Answer: d). Answers b) and c) are equivalent. However, lim n A n = A n for all n. 4. Sketch the graph of an increasing function f) such that f ) and F ) = are both decreasing. Since f ) is decreasing, f) must be concave down. Since F ) = ft) dt ft) dt is decreasing, its derivative must be negative, so f) must be negative. So for, f) must be negative, increasing and concave down. Here s one possible graph:
y = 6 " $! if > # % + OVER for Practice Problems SOLUTIONS
Practice Problems SOLUTIONS Problems 5-5) NOTE: You may use the following three summation formulae whenever needed. These formulae will also be given to you on the eam. [ ] nn + ) nn + )n + ) nn + ) ) i =, ) i =, 3) i 3 = 6 5. In each of the following, find f ). a) f) = e sin 3) f ) = e sin 3) + e sin 3) b) f) = ln arctan ) f ) = arctan arctan ) 3) 3 + 6. Find the general antiderivative of each of the following functions f): a) f) = 3 + 4 b) f) = sec tan 7. Find the following limits: a) lim b) tan 5) e e +, so F ) = 3 ln 3 + 4 tan + C., so F ) = sec sin + C. is an indeterminate form of type, so use l Hôpital s rule: lim lim tan = tan 5) e e lim tan = lim ) 5 + 5 e + e = 5. = π ) = π. 8. Do problems and 6 on pages 34 of the tet. In problem 6, use right hand endpoints for your estimate. Problem : The subintervals are [,.5], [.5, ], [,.5], [.5, ], [,.5], [.5, 3], and =.5. a) The speed is increasing over [, 3] so use left endpoints to get a lower estimate: ).5) + 6.).5) +.8).5) + 4.9).5) + 8.).5) + 9.4).5) = 34.7 feet. b) The speed is increasing over [, 3], so use right endpoints to get an upper estimate: 6.).5) +.8).5) + 4.9).5) + 8.).5) + 9.4).5) +.).5) = 44.8 feet.
Problem 6: Use 6 rectangles. The car s velocity is measured in km/hr, so convert the units on the t-ais from seconds to hours. So t = 5 =. Then the distance 36 7 is approimately 5 + 8 + 95 + + 5 + = 57 7 7 7 7 7 7 7.7967 km Note: The height of each rectangle is an estimate. Your estimates may differ slightly from those given above. 9. Let f) be continous on [a, b]. Consider the Riemann sum R n = f i ). a) Eplain why equals b a. n b a is the length of the interval [a, b]. We divide [a, b] into n subintervals of equal length, so each subinterval has length b a n b) Eplain why i equals a + i., so = b a n. We divide the interval [a, b] into n subintervals of width. See picture below.) a y =!. = if < y <!.4) = if < y <!.4) 3 i i n n }{{} y =!. if.99 < < ) ) Since i is the right endpoint of the ith subinterval, < y <!. ) if.99 < < ) i = a + } + {{ + + } = a + i. = i times y = c) Suppose f) on [a, b]. What does the epression f i ) represent? If f) on [a, b], then f i ) represents the area of the ith rectangle using right endpoints for the heights of the rectangles). The reason: the base of the ith rectangle is the subinterval.6[ i, i ], which has length. The height of the ith rectangle is fthe right endpoint) = f i ). b i, f i )) f) i i d) Suppose f) on [a, b]. What does the epression f i ) represent? If f) on [a, b], then f i ) represents the the opposite of the area of the ith rectangle using right endpoints for the heights of the rectangles).
y =.! = if. )! < y < ) = if. )! < y < ) y =. )! if.99 < < ). )! < y < if.99 < < ) y =! = if! 3 < <!.98) y = y = if! < <.) i! + 6 if i < < 3) 3. Let fr) be the function drawn below. Estimate y =! 3! 3 i, f i )) sum R 6. Then use geometry to find the eact value of 3 f) fr) dr by computing the Riemann 3 fr) dr. r R 6 = 6 fr i ) r. Note that r = 3 3) 6 =. Therefore R 6 = f ) + f ) + f) + f) + f) + f3) = ) + ) + + + + =. Using geometry to find the eact value of 3 3 fr) dr gives fr) dr = ) ) + ) ) + + 3)) =.5.
3. Is the integral e ln d positive, negative or zero? Eplain without integrating). Note that ln < for all values of in, ). Therefore ln d is negative. e 3. Evaluate the integral 3 ) d using the definition of the definite integral as a limit of a Riemann sum. Do NOT use techniques from sections 5.3 or 5.4.) Solution: Since a = and b = 3, = 3 = 3 and then n n i = + i = i 3 = 3i. n n So we get the Riemann sum n ) 3i n )3 3. Simplifying gives ) 3 7i3 = n n n 3 3 3 7i 3 = 3 8 i 3 = 3 n 8 n n+) = 3 8n+). Now take the n n n 3 n n 4 n n 4 4 4n limit of this last epression as n, getting 3 8 = 69. So 3 ) d = 69. 4 4 4 33. Evaluate the following definite integrals: a) 4 4 d y =! < y <! if! < < 4) d is the area under the two triangles in the following graph: e b) Their total area is ) + 3 3 ) = + 4.5 = 6.5. 4 + 5) d 4 + 5) d = 4 d + 5 d. 4 d is the area of a quarter circle of radius, so 4 d = 4 π) = π. Since π +. 34. Do problem 9 on page 374. 5 d = 5 ) =, we get that 4 + 5) d =
Note that f) = t t a t dt is an antiderivative of. We want f) =, so let a =. Therefore, f) = t dt. ) 35. Find d 3 e t dt. Hint: Look at problem #7 on page 373. d 3 3 Rewrite e t dt as e t dt + e t dt = e t dt + e t dt. Then ) d 3 e t dt = d ) 3 e t dt + e t dt = e 4 ) + e 6 3 ) d d 36. Let F ) = at =. Solution: F ) = = 3 e 6 e 4. 3 cosπt ) dt. Find the equation of the line tangent to the graph of F ) cosπt ) dt. F ) = cosπ ) by the st FTC. So F ) = cos π =, so the slope of the line is. F ) = cosπt ) dt =, so the point of tangency is, ). So the equation of the line is y = ), which can also be written as y = +. 37. Let g) = cos π t 4 + dt. Is g) is increasing or decreasing at = 3π? Why? Use the st FTC and the chain rule to get g ) = cos 4 sin ). + Then g 3π) = 4 + )) =. Since g 3π) is positive, g) is increasing at = 3π. 38. Let g) = te t e t ) dt for. On what intervals) is g) concave up? g ) = e e, so g ) = e + e e = e e. Set g ) =, getting e ) = =. Then do a sign analysis of g ): e + + + + + + + + + e ) + + + Since g ) > on, + ), g is concave up on, + ).
39. Let f be a continuous function. Determine whether each of the following statements is true or false. If a statement is true, briefly eplain why. If a statement is false, eplain why it s false or else give an eample that shows it s false. a) If f) d = 6, then 4 f) d =. < y < 6 3 4 False. As an eample, consider the function f) shown below. For this function, f) d = 6, but y = 6! 3! if < < ) 6! 3 < y < if < < 4) f) d =. b) If f) d = 6 and g) = f), then True by the following property of integrals: If c is a constant, then So g) d = c) If a b, then d) b a f) d. b a cf) d = c f) d = False. As an eample, consider the integral 4 g) d =. b a f) d. f) d = 6 =. shown in part a). Note that a b since 4. However,. f) d f) d, where f) is the function 4 f) d. f) d does equal False. As an eample, look again at the function f) shown in part a). There f) d = 6, but f) d = 6 3 = 4.5. So f) d < f) d.
e) y b a = = f) d = b a ft) dt. True. It doesn t matter what we call the independent variable in a particular function f. For eample, f) = 3 5 and ft) = t 3 5 represent the same y = function. So b 4. Evaluate the following: & ' $! if < < 5% " 5 # a f) d = b a ft) dt. a) sin 3 ) = π 3 b) tansin )) 6 To find tansin )), start by letting α = sin ). Note that α is in the 6 6 fourth quadrant. We get the following triangle: y = if < < 3.75) α = 3.75 if < y < 4 6 ) y =.75 +.5 if < < 5) Use the Pythagorean Theorem to get = 35. So tan α = 35, so tansin )) = 6 y =.9.9 35.! if < < 3.75) 4. Simplify csctan ). Your answer should not contain any trig or inverse trig functions.) Let θ = tan. First assume that >. Then we can draw the following triangle: θ The Pythagorean theorem tells us that the hypotenuse has length +. So csc θ = sin θ = +. Now assume that <. Then θ has negative measure, so our triangle lies in the fourth + quadrant; once again the Pythagorean Theorem tells us that csc θ =. Finally, note that if = then csctan ) = csc, which is undefined. So we can conclude the following: + csctan ) = for all.
4. Suppose that f ) = 5 ) and f ) = 5π. Find f). 6 Since f ) = 5 ) = 5 we see that f) = 5 sin + C. Since f ) = 5π, we have: 6 So f) = 5 sin + π 3. 43. Evaluate the following: a) b) 9 cosπi) i=5 5 sin ) + C = 5π 6 5 π ) + C = 5π 6 6 C = π 6. 9 cosπi) = cos5π) + cos6π) + cos7π) + cos8π) + cos9π) i=5 5 4n = ) + + ) + + ) =. Note that n is a constant in this sum. So 44. Approimate the integral using L 3. Note that = π. Then π 5 4n = 5 4n = 5 4n n = 5 4. cos d using R 3. Then approimate the same integral R 3 = f π) π + fπ) π + f3π) π π ) = cos π + cosπ) π + cos = π + ) π + π = π ) 3π π and L 3 = f) π + fπ) π + fπ) π = cos) π π ) + cos = π + π + ) π =. π + cosπ) π
45. Evaluate the integral + 4) d using the definition of the definite integral as a limit of a Riemann sum. Do NOT use techniques from sections 5.3 or 5.4.) Since a = and b =, = =. Moreover, n n i = + i = + i = + i. n n So we get the Riemann sum ) ) + i n + 4 + i ) = ) ) n n n + i n + 4 + i ) ) = n n + i + i + 4 + 4i n n n = 5 n = n + 6 i + n n 3 5 + 6i n + i n ) = n 5 + n 6i n + n i n i = 5 n + 6 nn+) + nn+)n+) n n n 3 6 = 5 + 3 n+ n + n3 +3n +n 6n 3. Now take the limit of this last epression as n, getting 5 + 3 + 3 = 5 3. So + ) d = 5 3. 46. Epress the following limit as a definite integral, using a =. Then epress the same limit as a different integral, this time using a =. + 3i n )5 3 n lim n To epress the limit as a definite integral, we must put it in the form we must find a, b and f). b a f) d, so a) First let a =. It looks like = 3 b a. Therefore, since = and a =, b n n must equal 5. Since i = a + i, we know that i = + 3i. The function in the n integrand is + 3i n )5, so f i ) = + i ) 5. So f) = + ) 5. 5 Hence the integral is + )5 d. b) Now let a =. Again, it looks like = 3 b a. Therefore, since = and a =, n n b must equal 3. i = a + i, so here i = + 3i = 3i. Again the function in the n n integrand is + 3i n )5, so f i ) = + i ) 5. So f) = + 5. Hence the integral is + 5 d. 47. The graph of a function f) over the interval [ 5, 5] is shown below. Suppose that left region has area A, the middle region has area B, and the right region has area C. Suppose we know that A = 5.6 and B =.3. a) What is 5 f) d?
5 f)d = A + B = 5.6 +.3 = 4.3 b) Suppose that we know that 5 5 + 5) ) + )! )! 5) < y < if! 5 < <! ) 8 + 5) ) + )! )! 5) 8 5 5 f) d = 7.. What is C? < y < if < < 5) f)d = 7. A + B C = 7. 4.3 C = 7. C =.8. 48. Evaluate the following definite integrals: a) b) 3 3 4 9 d 4 9 d = 4 9 d. The equation y = 9 represents the 3 9 d = π3 ) = semi-circle with radius 3, centered at the origin. So 9π. So 3 + ) d 4 9 d = 4 9π = 8π. + ) d. There are several ways to compute this integral. One is to rewrite 3 the integral as yd = + d. Use geometry to find < y < if! < < 3) getting 3. For d, we get 3 )) =. So d see below), +) d = 3 + = 33. Another way to do the problem is to graph the function f) = + see below) and use geometry to find the area. The result is again 33.
y = + < y < + if! < < 3) 49. Let f) = sin 3 + ). Find a function F ) such that F ) = f) and F ) = 7. Note that F ) = F ) =, not 7. So let F ) = 7 + conditions. 5. Do problems 39 and 4 on page 46. Problem 39: F ) = sint 3 + ) dt is an antiderivative of sin 3 + ). + 3. However, sint 3 + ) dt. This satisfies all the necessary Problem 4: Use the st FTC and the chain rule: g ) = sin + sin 4 cos.