We know that if f is a continuous nonnegative function on the interval [a, b], then b


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1 1 Ares Between Curves c 22 Donld Kreider nd Dwight Lhr We know tht if f is continuous nonnegtive function on the intervl [, b], then f(x) dx is the re under the grph of f nd bove the intervl. We re going to extend this notion bit by considering how to find the re between two functions. To be specific, suppose we re given two continuous functions, f top nd g bottom defined on the intervl [, b], with g bottom (x) f top (x) for ll x in the intervl. How do we find the re bounded by the two functions over tht intervl? We hve used the nottion f top nd g bottom for obvious resons. However, we wnt to cution you tht ll of the subsequent nlysis relly does ssume tht f lies bove or is equl to g t every point throughout the intervl. So, you wnt to be sure in problemsolving tht you hve verified tht this is the cse before using the formul tht we will develop next. The re of the region between the two curves nd bove the intervl [, b] equls the re of the region under the grph of f top on tht intervl minus the re of the region under the grph of g bottom on the sme intervl. Thus, the re of the region between the two curves equls f top (x) dx g bottom (x) dx = (f top (x) g bottom (x)) dx The lst integrl is normlly the form in which we express the re between the two curves. But remember tht to pply the formul, we hve to know which curve is on top nd which is on bottom, nd we hve to be certin tht tht reltionship is mintined throughout the intervl. Note though tht the verticl plcement of the curves does not mtter. For exmple, both of them could lie below the xxis, or one bove nd one below, nd the formul would still hold. Exmple 1: Find the re of the region between the grphs of y = x 2 nd y = x 3 for x 1. Solving for the point(s) of intersection, we find tht the curves intersect t x =, 1: x 2 = x 3 x 2 (x 1) = implies x =, x = 1. Here is wht quick sketch by hnd might look like:
2 2 To determine which curve is on top, we plug in convenient vlue of x to find tht y = x 2 is on top throughout the intervl [, 1]. Thus, we cn use the formul bove: The re of the region between the two curves equls 1 ( x 2 x 3) dx = ( x 3 /3 x 4 /4 ) 1 = 1/3 1/4 = 1/12. Exmple 2: Find the re of the region between y = e x nd y = 1 1+x on the intervl [, 1]. The grph is shown. Becuse the curves intersect t x =, nd y = e x is incresing while y = 1/(1 + x) is decresing, we know tht e x is on top. Thus, the re of the region between the curves equls 1 ( e x 1 ) dx = (e x ln 1 + x ) x = e ln 2 e + ln 1 = e ln 2 1 Exmple 3: Find the re of the region bounded by y = x 2 2x nd y = 4 x 2. To solve this problem, we need sketch so tht we cn determine which function is on top over which intervls. We will begin by determining the points of intersection. x 2 2x = 4 x 2 2x 2 2x 4 = 2(x 2 x 2) = 2(x 2)(x + 1) = So, x = 2 or x =. Both functions re qudrtic polynomils, so their grphs re prbols, one opening up nd the other down. We should recognize from the sign on x 2 which curve is on top, or we cn test vlue of x to find out.
3 3 Thus, the re of the region cn be gotten by pplying our integrl formul to obtin: 2 ) (4 x 2 (x 2 2x)) dx = (4x 2x x2 = 8 16 ( ) = 9 Exmple 4: Find the re of the region bounded by the two curves y = x 3 9x nd y = 9 x 2. [Hint: You probbly will need to know tht x+1 is fctor of x 3 +x 2 9x 9.] Let s find the points of intersection: x 3 9x = 9 x 2 x 3 + x 2 9x 9 = (x + 1)(x 2 9) = (x + 1)(x 3)(x + 3) = Note tht to obtin the next to the lst eqution bove, we used the hint nd divided x 3 + x 2 9x 9 by x + 1. A sketch of the grphs follows: Thus, the re cn be gotten by pplying our integrl formul twice once to the intervl [ 3, ] where x 3 9x is on top, nd once to the intervl [, 3] where 9 x 2 is on top nd dding the results together: (x 3 9x (9 x 2 )) dx (9 x 2 (x 3 9x)) dx Routine clcultion gives the nswer Try it for yourself nd verify this result. Exmple 5: Find the re between sin x nd cos x on [, π/4]. Here is sketch:
4 4 So, the re is π/4 (cos x sin x) dx = (sin x + cos x) π/4 = sin(π/4) + cos(π/4) (sin + cos ) = ( + 1) 2 2 = 2 1 Functions of y: Thus fr, we hve only considered functions of x. We could just s well consider two functions of y, sy, x = f Left (y) nd x = g Right (y) defined on the intervl [c, d] on the yxis s in the sketch below: Then the re between the grphs cn be found by subdividing the intervl [c, d] on the yxis, nd using horizontl rectngulr re elements. In tht cse, we get tht the re between the two curves is the definite integrl d c (g Right (y) f Left (y)) dy Exmple 6: Find the re under the grph of y = ln x nd bove the intervl [1, e] on the xxis. We know tht integrting with respect to x yields the definite integrl e ln x dx. However, suppose we 1 do not know (or remember, or wnt to investigte) n ntiderivtive for ln x. Then we cn try to solve this problem by integrting insted with respect to y. The functions re x = e on the right, x = e y on the left, over the intervl from y = to y = 1: 1 (e e y ) dy = (ey e y ) 1 = e e + 1 = 1
5 5 Thus, our problem is solved. Alterntively, we could go bck to where we begn. It turns out tht by ppliction of the Integrtion by Prts formul to the integrl ln x dx, x ln x x is n ntiderivtive of ln x. (Just clculte the derivtive to verify this result.) Thus, we cn evlute the originl integrl with respect to x: e 1 ln x dx = (x ln x x) e 1 = e e ( 1) = 1 As we should, we get the sme nswer for the re of the region integrting with respect to either x or y. Exercises: Problems Check wht you hve lerned! Videos: Tutoril Solutions See problems worked out!
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