MA13, Chapter 8: Idea of the Integral (pp. 155-187, Gootman) Chapter Goals: Understand the relationship between the area under a curve and the definite integral. Understand the relationship between velocit (speed), distance and the definite integral. Estimate the value of a definite integral. Understand the summation, or Σ, notation. Understand the formal definition of the definite integral. Assignments: Assignment 18 Assignment 19 The basic idea: The first two problems are eas to solve as certain problem ingredients are constant. Eample 1 (Eas area problem): Find the area of the region in the plane bounded above b the graph of the function f() =, below b the -ais, on the left b the line = 1, and on the right b the line = 5. 0 1 5 Eample (Eas distance traveled problem): Suppose a car is traveling due east at a constant velocit of 55 miles per hour. How far does the car travel between noon and :00 pm? General philosoph: B means of the integral, problems similar to the previous ones can be solved when the ingredients of the problem are variable. In this Chapter, we learn how to estimate a solution to these more comple problems. The ke idea is to notice that the value of the function does not var ver much over a small interval, and so it is approimatel constant over a small interval. B the end of Chapter 9 we will be able to solve these problems eactl, and b the end of Chapter 10 we will be able to solve them both eactl and easil. Eample 3: Estimate the area under the graph of = + 1 for between 0 and in two different was: 5 (a) Subdivide the interval [0, ] into four equal subintervals and use the left endpoint of each subinterval as sample point. 3 (b) Subdivide the interval [0, ] into four equal subintervals and use the right endpoint of each subinterval as sample point. 1.5 0.5 0 1 5 3 Find the difference between the two estimates (right endpoint estimate minus left endpoint estimate). 85 1.5 0.5 0 1
9 Eample 4: Estimate the area under the graph of = 3 for between 0 and. Use a partition that consists of four equal subintervals of [0,] and use the left endpoint of each subinterval as a sample point. 5.196 3 1.73 1 0 1 Note: In the previous two eamples we sstematicall chose the value of the function at one of the endpoints of each subinterval. However, since the guiding idea is that we are assuming that the values of the function over a small subinterval do not change b ver much, then we could take the value of the function at an point of the subinterval as a good sample or representative value of the function. We could also have chosen small subintervals of different lengths. However, we are tring to establish a sstematic procedure that works well in general. Getting better estimates: We can onl epect the previous answers to be approimations of the correct answers. This is because the values of the function do change on each subinterval, even though the do not change b much. If, however, we replace the subintervals we used b smaller subintervals we can reasonabl epect the values of the function to var b much less on each thinner subinterval. Thus, we can reasonabl epect that the area of each thinner vertical strip under the graph of the function to be more accuratel approimated b the area of these thinner rectangles. Then if we add up the areas of all these thinner rectangles, we should get a much more accurate estimate for the area of the original region. Here is Eample 3(b), revisited: = + 1 on [0,] n = 4 equal subintervals Area 5 = + 1 on [0,] n = 8 equal subintervals Area 4.315 0 1 0 1 We will see later that the eact value of the area under consideration in Eample 3 is 11 3 3.66. 86
Eample 5: Estimate the area of the ellipse given b the equation 4 + = 49 as follows: The area of the ellipse is 4 times the area of the part of the ellipse in the first quadrant ( and positive). Estimate the area of the ellipse in the first quadrant b solving for in terms of. Estimate the area under the graph of b dividing the interval [0,3.5] into four equal subintervals and using the left endpoint of each subinterval. 0 The area of the ellipse (using the above method) is approimatel Trapezoids versus rectangles: We could use trapezoids instead of rectangles to obtain better estimates, even though the calculations get a little bit more complicated. This will occur in some of the latter eamples. We recall that the area of a trapezoid is Area of a trapezoid = (h 1 +h ) b. h b h h 1 h 1 b Eample 6: A train travels in a straight westward direction along a track. The velocit of the train varies, but it is measured at regular time intervals of 1/10 hour. The measurements for the first half hour are time 0 0.1 0. 0.3 0.4 0.5 velocit 0 10 15 18 0 5 We will see later that the total distance traveled b the train is equal to the area underneath the graph of the velocit function and ling above the t-ais. Compute the total distance traveled b the train during the first half hour b assuming the velocit is a linear function of t on the subintervals. (The velocit in the table is given in miles per hour.) v 5 0 18 15 10 0 0.1 0. 0.3 0.4 0.5 t 87
Eample 7: Estimate the area under the graph of = 1 for between 1 and 31 in two different was: (a) Subdivide the interval [1, 31] into 30 equal subintervals and use the left endpoint of each subinterval as sample point. (b) Subdivide the interval [1, 31] into 30 equal subintervals and use the right endpoint of each subinterval as sample point. Find the difference between the two estimates (left endpoint estimate minus right endpoint estimate). Sigma (Σ) notation: In approimating areas we have encountered sums with man terms. A convenient wa of writing such sums uses the Greek letter Σ (which corresponds to our capital S) and is called sigma notation. More precisel, if a 1,a,...,a n are real numbers we denote the sum a 1 +a + +a n b using the notation a k. The integer k is called an inde or counter and takes on (in this case) the values 1,,...,n. For eample, whereas 6 k = 1 + +3 +4 +5 +6 = 1+4+9+16+5+36 = 91 6 k = 3 +4 +5 +6 = 9+16+5+36 = 86. k=3 Eample 8: Evaluate the sum 5 (k 1). 88
Eample 9: Evaluate the sum 6 (6k 3 +3). k= Eample 10: Evaluate the sum 5 (3k +k). Eample 11: Evaluate the sum 11 75. Eample 1: Evaluate the sum 73 5 3. 89
The idea we have used so far is to break up or subdivide the given interval [a,b] into lots of little pieces, or subintervals, on each of which the variable, and thus the function f(), does not change much. The technical phrase for doing this is to partition [a,b]. Definition of a partition: Apartition ofaninterval[a,b] isacollection ofpoints{ 0, 1,,..., n 1, n }, listed increasingl, on the -ais with a = 0 and n = b. That is: a = 0 < 1 < <... < n 1 < n = b. These points subdivide the interval [a,b] into n subintervals: [a, 1 ], [ 1, ], [, 3 ],..., [ n 1,b]. The k-th subinterval is thus of the form [ k 1, k ] and it has length k = k k 1. Assumption: Set P = ma { i}. We will alwas assume that our partition P is such that P 0 1 i n as n. In other words, we alwas assume that the length of the longest (and as a consequence of all) subinterval(s) tend(s) to zero whenever the number of subintervals in our partition P becomes ver large. The definite integral: Let f() be a function defined on an interval [a,b]. Partition the interval [a,b] in n subintervals of lengths 1,..., n,respectivel. Fork = 1,...,npickarepresentativepointp k inthecorrespondingk-thsubinterval. The definite integral of the function f from a to b is defined as ) lim f(p k ) k = lim (f(p 1 ) 1 +f(p ) + +f(p n ) n = lim n n P 0 and it is denoted b b a f()d. f(p k ) k The sum f(p k ) k is called a Riemann sum in honor of the German mathematician Bernhard Riemann (186-1866), who developed the above ideas in full generalit. The smbol is called the integral sign. It is an elongated capital S, of the kind used in the 1600s and 1700s. The letter S stands for the summation performed in computing a Riemann sum. The numbers a and b are called the lower and upper limits of integration, respectivel. The function f() is called the integrand and the smbol d is called the differential of. You can think of the d as representing what happens to the term in the limit, as the size of the subintervals gets closer and closer to 0. Note: The role of in a definite integral is the one of a dumm variable. In fact have the same meaning. The represent the same number. b a d and b a t dt Note: We recall from Chapter 3 that a limit does not necessaril eist. However: Theorem: Let f() be a continuous function on the interval [a,b] then limit used in the definition of the definite integral eists. b a f()d eists. That is, the Regular partitions: As we observed earlier, it is computationall easier to partition the interval [a, b] into n subintervals of equal length. Therefore each subinterval has length = b a (we drop the inde k as it is n no longer necessar). In this case, there is a simple formula for the points of the partition, namel: 0 = a, 1 = a+, = a+,... k = a+k,..., n 1 = a+(n 1), n = b or, more concisel, k = a+k b a n for k = 0,1,,...,n. 90
Right versus left endpoint estimates: Observe that k, the right endpoint of the k-th subinterval, is also the left endpoint of the (k+1)-th subinterval. Thus the Riemann sum estimate for the definite integral of a function f defined over aninterval [a,b] can bewritten ineither of thefollowing two forms n 1 f( k ) k+1 k=0 f( k ) k 0 1 Left endpoint Riemann sum estimate depending on whether we use left or right endpoints, respectivel. If we are dealing with a regular partition, the above sums become n 1 f(a+k ) k=0 f(a+k ) respectivel, with = (b a)/n and k = a+k for k = 0,1,,...,n. 0 1 Right endpoint Riemann sum estimate Eample 13: Suppose ou estimate the integral 7 1 8 1+k. 8 d b evaluating the sum If ou use =., what value should ou use for n, the upper limit of the summation? Eample 14: Suppose ou estimate the integral 10 (+k ). d b evaluating the sum If ou use n = 10 intervals, what value should ou use for, the length of each interval? Eample 15: Suppose ou estimate the integral 0 6 d b the sum [A+B(k )+C(k ) ], where n = 30 and = 0.. The terms in the sum equal areas of rectangles obtained b using right end points of the subintervals of length as sample points. What is the value of B? 91
Eample 16: Suppose ou estimate the integral 15 5 3 d b the sum (a+k ) 3, where n = 50 and = 0.. The terms in the sum equal areas of rectangles obtained b using right end points of the subintervals of length as sample points. What is the value of a? Eample 17: Suppose ou estimate the integral 15 3 f()d b adding the areas of n rectangles of equal base length, and ou use the right endpoint of each subinterval to determine the height of each rectangle. If the sum ou evaluate is written as what is A? f(3+k A/n) A/n, Eample 18: Suppose ou estimate the integral 9 3 f(3+k ). f()d b evaluating a sum If ou use n = 6 intervals of equal length, what value should ou use for? Eample 19: Suppose ou estimate the area under the graph of f() = 3 from = 4 to = 4 b adding the areas of rectangles as follows: partition the interval into 0 equal subintervals and use the right endpoint of each interval to determine the height of the rectangle. What is the area of the 15 th rectangle? 9
Eample 0: Supposeou estimate the area under the graph of f() = 1 from = 1 to = 11 b adding the areas of rectangles as follows: partition the interval into 50 equal subintervals and use the left endpoint of each interval to determine the height of the rectangle. What is the area of the 4 th rectangle? Eample 1: Suppose ou are given the following data points for a function f(): 1 3 4 f() 5 8 1 If f is a linear function on each interval between the given points, find 4 1 f()d. Eample : Suppose f() is the greatest integer function, i.e., f() equals the greatest integer less than or equal to. So for eample f(.3) =, f(4) = 4, and f(6.9) = 6. Find 10 6 f()d. (Hint: Draw a picture. See also eample 18 in Chapter 1 and eample 19 in Chapter 3.) 93
94