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376 Chapter 5: Integration Substitution and Area Between Curves There are two methods for evaluating a definite integral by substitution. The first method is to find an antiderivative using substitution, and then to evaluate the definite integral by applying the Fundamental Theorem. We used this method in Examples 8 and 9 of the pre- ceding section. The second method extends the process of substitution directly to definite integrals. We apply the new formula introduced here to the problem of computing the area between two curves. Substitution Formula In the following formula, the limits of integration change when the variable of integration is changed by substitution. 5.6 THEOREM 6 Substitution in Definite Integrals If is continuous on the interval [a, b] and ƒ is continuous on the range of g, then L b a ƒsgsxdd # g¿sxd dx = L gsbd gsad ƒsud du g¿ 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 376 Proof Let F denote any antiderivative of ƒ. Then, To use the formula, make the same u-substitution and you would use to evaluate the corresponding indefinite integral. Then integrate the trans- formed integral with respect to u from the value g(a) (the value of u at ) to the value g(b) (the value of u at ). EXAMPLE 1 Substitution by Two Methods Evaluate Solution We have two choices. Method 1: Transform the integral and evaluate the transformed integral with the trans- formed limits given in Theorem 6. Evaluate the new definite integral. Method 2: Transform the integral as an indefinite integral, integrate, change back to x, and use the original x-limits. = 2 3 c2 3>2 - 0 3>2 d = 2 3 c222d = 422 3 = 2 3 css1d 3 + 1d 3>2 - ss-1d 3 + 1d 3>2 d L 1 -1 3x 2 2x 3 + 1 dx = 2 3 sx 3 + 1d 3>2 d -1 1 = 2 3 sx 3 + 1d 3>2 + C = 2 3 u 3>2 + C L 3x 2 2x 3 + 1 dx = L 1u du = 2 3 c2 3>2 - 0 3>2 d = 2 3 c222d = 422 3 = 2 3 u 3>2 d 0 2 = L 2 0 1u du L 1 -1 3x 2 2x 3 + 1 dx L 1 -1 3x 2 2x 3 + 1 dx. x = b x = a du = g¿sxd dxu = gsxd = L gsbd gsad ƒsud du. = Fsudd u = gsad u = gsbd = Fsgsbdd - Fsgsadd L b a ƒsgsxdd # g¿sxd dx = Fsgsxddd x = a x = b 5.6 Substitution and Area Between Curves 377 = ƒsgsxddg¿sxd = F¿sgsxddg¿sxd d dx Fsgsxdd Fundamental Theorem, Part 2 Let When When x = 1, u = s1d 3 + 1 = 2. x =-1, u = s -1d 3 + 1 = 0. u = x 3 + 1, du = 3x 2 dx. Let u = x 3 + 1, du = 3x 2 dx. Integrate with respect to u. Replace u by x 3 + 1. Use the integral just found, with limits of integration for x. 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 377 Which method is better—evaluating the transformed definite integral with trans- formed limits using Theorem 6, or transforming the integral, integrating, and transforming back to use the original limits of integration? In Example 1, the first method seems easier, but that is not always the case. Generally, it is best to know both methods and to use whichever one seems better at the time. EXAMPLE 2 Using the Substitution Formula Definite Integrals of Symmetric Functions The Substitution Formula in Theorem 6 simplifies the calculation of definite integrals of even and odd functions (Section 1.4) over a symmetric interval (Figure 5.26).[-a, a] =-c s0d 2 2 - s1d 2 2 d = 1 2 =-c u 2 2 d 1 0 =- L 0 1 u du L p>2 p>4 cot u csc 2 u du = L 0 1 u # s -dud 378 Chapter 5: Integration x y 0 a–a (b) x y 0 a–a (a) FIGURE 5.26 (a) ƒ even, (b) ƒ odd, 1 a -a ƒsxd dx = 0 1 a -a ƒsxd dx = 2 1 a 0 ƒsxd dx Theorem 7 Let ƒ be continuous on the symmetric interval (a) If ƒ is even, then (b) If ƒ is odd, then L a -a ƒ(x) dx = 0. L a -a ƒsxd dx = 2 L a 0 ƒsxd dx. [-a, a]. Let When When u = p>2, u = cot (p>2) = 0. u = p>4, u = cot (p>4) = 1. - du = csc 2 u du. u = cot u, du =-csc 2 u du, 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 378 Proof of Part (a) The proof of part (b) is entirely similar and you are asked to give it in Exercise 86. The assertions of Theorem 7 remain true when ƒ is an integrable function (rather than having the stronger property of being continuous), but the proof is somewhat more diffi- cult and best left to a more advanced course. EXAMPLE 3 Integral of an Even Function Evaluate Solution Since satisfies it is even on the symmet- ric interval so Areas Between Curves Suppose we want to find the area of a region that is bounded above by the curve below by the curve and on the left and right by the lines and (Figure 5.27). The region might accidentally have a shape whose area we could find with geometry, but if ƒ and g are arbitrary continuous functions, we usually have to find the area with an integral. To see what the integral should be, we first approximate the region with n vertical rec- tangles based on a partition of [a, b] (Figure 5.28). The area of the kth rectangle (Figure 5.29) is ¢A k = height * width = [ƒsc k d - gsc k d] ¢x k . P = 5x 0 , x 1 , Á , x n 6 x = bx = ay = gsxd, y = ƒsxd, = 2 a 32 5 - 32 3 + 12b = 232 15 . = 2 c x 5 5 - 4 3 x 3 + 6xd 0 2 L 2 -2 sx 4 - 4x 2 + 6d dx = 2 L 2 0 sx 4 - 4x 2 + 6d dx [-2, 2], ƒs -xd = ƒsxd,ƒsxd = x 4 - 4x 2 + 6 L 2 -2 sx 4 - 4x 2 + 6d dx. = 2 L a 0 ƒsxd dx = L a 0 ƒsud du + L a 0 ƒsxd dx = L a 0 ƒs -ud du + L a 0 ƒsxd dx =- L a 0 ƒs -uds -dud + L a 0 ƒsxd dx =- L -a 0 ƒsxd dx + L a 0 ƒsxd dx L a -a ƒsxd dx = L 0 -a ƒsxd dx + L a 0 ƒsxd dx 5.6 Substitution and Area Between Curves 379 Additivity Rule for Definite Integrals Order of Integration Rule Let When When x =-a, u = a. x = 0, u = 0. u =-x, du =-dx. ƒ is even, so ƒs -ud = ƒsud. x y a b Lower curve y H11005 g(x) Upper curve y H11005 f(x) FIGURE 5.27 The region between the curves and and the lines and x = b.x = a y = gsxdy = ƒsxd x y y H11005 f(x) y H11005 g(x) b H11005 x n x nH110021 a H11005 x 0 x 1 x 2 FIGURE 5.28 We approximate the region with rectangles perpendicular to the x-axis. x y a b (c k , f(c k )) f(c k ) H11002 g(c k ) H9004A k c k (c k , g(c k )) H9004x k FIGURE 5.29 The area of the kth rectangle is the product of its height, and its width, ¢x k .ƒsc k d - gsc k d, ¢A k 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 379 380 Chapter 5: Integration We then approximate the area of the region by adding the areas of the n rectangles: Riemann Sum As the sums on the right approach the limit because ƒ and g are continuous. We take the area of the region to be the value of this integral. That is, A = lim ƒƒ P ƒƒ :0 a n k = 1 [ƒsc k d - gsc k d] ¢x k = L b a [ƒsxd - gsxd] dx. 1 b a [ƒsxd - gsxd] dx7P7 : 0, A L a n k = 1 ¢A k = a n k = 1 [ƒsc k d - gsc k d] ¢x k . DEFINITION Area Between Curves If ƒ and g are continuous with throughout [a, b], then the area of the region between the curves and from a to b is the inte- gral of from a to b: A = L b a [ƒsxd - gsxd] dx. ( f - g) y = gsxdy = fsxd ƒsxd Ú gsxd When applying this definition it is helpful to graph the curves. The graph reveals which curve is the upper curve ƒ and which is the lower curve g. It also helps you find the limits of integration if they are not already known. You may need to find where the curves inter- sect to determine the limits of integration, and this may involve solving the equation for values of x. Then you can integrate the function for the area be- tween the intersections. EXAMPLE 4 Area Between Intersecting Curves Find the area of the region enclosed by the parabola and the line Solution First we sketch the two curves (Figure 5.30). The limits of integration are found by solving and simultaneously for x. Equate ƒ(x) and g(x). Rewrite. Factor. Solve. The region runs from to The limits of integration are The area between the curves is = a4 + 4 2 - 8 3 b - a-2 + 1 2 + 1 3 b = 9 2 = L 2 -1 s2 + x - x 2 d dx = c2x + x 2 2 - x 3 3 d -1 2 A = L b a [ƒsxd - gsxd] dx = L 2 -1 [s2 - x 2 d - s -xd] dx a =-1, b = 2.x = 2.x =-1 x =-1, x = 2. sx + 1dsx - 2d = 0 x 2 - x - 2 = 0 2 - x 2 =-x y =-xy = 2 - x 2 y =-x.y = 2 - x 2 ƒ - gƒsxd = gsxd x y 0–1 1 2 (–1, 1) (x, f(x)) y H11005 2 H11002 x 2 (x, g(x)) H9004x y H11005 –x (2, –2) FIGURE 5.30 The region in Example 4 with a typical approximating rectangle. 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 380 If the formula for a bounding curve changes at one or more points, we subdivide the re- gion into subregions that correspond to the formula changes and apply the formula for the area between curves to each subregion. EXAMPLE 5 Changing the Integral to Match a Boundary Change Find the area of the region in the first quadrant that is bounded above by and be- low by the x-axis and the line Solution The sketch (Figure 5.31) shows that the region’s upper boundary is the graph of The lower boundary changes from for to for (there is agreement at ). We subdivide the region at into subre- gions A and B, shown in Figure 5.31. The limits of integration for region A are and The left-hand limit for re- gion B is To find the right-hand limit, we solve the equations and simultaneously for x: Equate ƒ(x) and g(x). Square both sides. Rewrite. Factor. Solve. Only the value satisfies the equation The value is an extrane- ous root introduced by squaring. The right-hand limit is We add the area of subregions A and B to find the total area: Integration with Respect to y If a region’s bounding curves are described by functions of y, the approximating rectangles are horizontal instead of vertical and the basic formula has y in place of x. = 2 3 s8d - 2 = 10 3 . = 2 3 s2d 3>2 - 0 + a 2 3 s4d 3>2 - 8 + 8b - a 2 3 s2d 3>2 - 2 + 4b = c 2 3 x 3>2 d 0 2 + c 2 3 x 3>2 - x 2 2 + 2xd 2 4 Total area = L 2 0 1x dx (')'* area of A + L 4 2 s1x - x + 2d dx ('''')''''* area of B For 2 … x … 4: ƒsxd - gsxd = 1x - sx - 2d = 1x - x + 2 For 0 … x … 2: ƒsxd - gsxd = 1x - 0 = 1x b = 4. x = 11x = x - 2.x = 4 x = 1, x = 4. sx - 1dsx - 4d = 0 x 2 - 5x + 4 = 0 x = sx - 2d 2 = x 2 - 4x + 4 1x = x - 2 y = x - 2 y = 1xa = 2. b = 2.a = 0 x = 2x = 22 … x … 4 gsxd = x - 20 … x … 2gsxd = 0ƒsxd = 1x. y = x - 2. y = 1x 5.6 Substitution and Area Between Curves 381 HISTORICAL BIOGRAPHY Richard Dedekind (1831–1916) x y 0 1 2 42 y H11005 H20857x y H11005 0 y H11005 x H11002 2 (x, f(x)) (x, f(x)) (x, g(x)) (x, g(x)) A B (4, 2) Area H11005 2 0 H20857x dx Area H11005 4 2 (H20857x H11002 x H11001 2) dx L L FIGURE 5.31 When the formula for a bounding curve changes, the area integral changes to become the sum of integrals to match, one integral for each of the shaded regions shown here for Example 5. 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 381 For regions like these use the formula In this equation ƒ always denotes the right-hand curve and g the left-hand curve, so is nonnegative. EXAMPLE 6 Find the area of the region in Example 5 by integrating with respect to y. Solution We f irst sketch the region and a typical horizontal rectangle based on a parti- tion of an interval of y-values (Figure 5.32). The region’s right-hand boundary is the line so The left-hand boundary is the curve so The lower limit of integration is We find the upper limit by solving and simultaneously for y: Rewrite. Factor. Solve. The upper limit of integration is (The value gives a point of intersection below the x-axis.) The area of the region is This is the result of Example 5, found with less work. = 4 + 4 2 - 8 3 = 10 3 . = c2y + y 2 2 - y 3 3 d 0 2 = L 2 0 [2 + y - y 2 ] dy A = L b a [ƒs yd - gs yd] dy = L 2 0 [y + 2 - y 2 ] dy y =-1b = 2. y =-1, y = 2 s y + 1ds y - 2d = 0 y 2 - y - 2 = 0 y + 2 = y 2 x = y 2 x = y + 2y = 0. gs yd = y 2 .x = y 2 ,ƒs yd = y + 2.x = y + 2, ƒs yd - gs yd A = L d c [ƒs yd - gs yd] dy. x H11005 f(y) y y x x x y x H11005 g(y) 0 c d x H11005 g(y) x H11005 f(y) 0 c d 0 c d x H11005 f(y) x H11005 g(y) 382 Chapter 5: Integration Equate and gsyd = y 2 . ƒsyd = y + 2 x y y H11005 02 40 1 2 (g(y), y) ( f(y), y) f(y) H11002 g(y) (4, 2) x H11005 y H11001 2 x H11005 y 2 H9004y FIGURE 5.32 It takes two integrations to find the area of this region if we integrate with respect to x. It takes only one if we integrate with respect to y (Example 6). 4100 AWL/Thomas_ch05p325-395 8/20/04 9:57 AM Page 382 Combining Integrals with Formulas from Geometry The fastest way to find an area may be to combine calculus and geometry. EXAMPLE 7 The Area of the Region in Example 5 Found the Fastest Way Find the area of the region in Example 5. Solution The area we want is the area between the curve and the x-axis, minus the area of a triangle with base 2 and height 2 (Figure 5.33): Conclusion from Examples 5–7 It is sometimes easier to find the area between two curves by integrating with respect to y instead of x. Also, it may help to combine geometry and calculus. After sketching the region, take a moment to think about the best way to proceed. = 2 3 s8d - 0 - 2 = 10 3 . = 2 3 x 3>2 d 0 4 - 2 Area = L 4 0 1x dx - 1 2 s2ds2d y = 1x, 0 … x … 4, 5.6 Substitution and Area Between Curves 383 02 2 4 1 2 Area H11005 2 (4, 2) x y y H11005 0 y H11005 x H11002 2 y H11005 H20857x 2 FIGURE 5.33 The area of the blue region is the area under the parabola minus the area of the triangle (Example 7). y = 1x 4100 AWL/Thomas_ch05p325-395 8/20/04 9:58 AM Page 383 Commercial_CD 4100 AWL/Thomas_ch05p325-395

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