APPLICATIONS OF INTEGRATION

Transcription

1 6 APPLICATIONS OF INTEGRATION The volume of a sphere is the limit of sums of volumes of approimating clinders. In this chapter we eplore some of the applications of the definite integral b using it to compute areas between curves, volumes of solids, and the work done b a varing force. The common theme is the following general method, which is similar to the one we used to find areas under curves:we break up a quantit Q into a large number of small parts. We net approimate each small part b a quantit of the form f i* and thus approimate Q b a Riemann sum. Then we take the limit and epress Q as an integral. Finall we evaluate the integral using the Fundamental Theorem of Calculus or the Midpoint Rule. 44

2 a S =ƒ = FIGURE S=s(,) a b, ƒd b 6. AREAS BETWEEN CURVES In Chapter 5 we defined and calculated areas of regions that lie under the graphs of functions. Here we use integrals to find areas of regions that lie between the graphs of two functions. Consider the region S that lies between two curves f and t and between the vertical lines a and b, where f and tare continuous functions and f t for all in a, b. (See Figure.) Just as we did for areas under curves in Section 5., we divide S into n strips of equal width and then we approimate the ith strip b a rectangle with base and height f i* t i*. (See Figure. If we like, we could take all of the sample points to be right endpoints, in which case i* i.) The Riemann sum n f i* t i* i is therefore an approimation to what we intuitivel think of as the area of S. f( i *) f( i *)-g( i *) FIGURE a _g(*) i This approimation appears to become better and better as nl. Therefore we define the area A of the region S as the limiting value of the sum of the areas of these approimating rectangles. Î i * (a) Tpical rectangle b a A nl lim n f i* t i* i (b) Approimating rectangles b. Therefore we have the fol- We recognize the limit in () as the definite integral of lowing formula for area. The area A of the region bounded b the curves f, t, and the lines a, b, where f and tare continuous and f t for all in a, b, is f t d A b a f t Notice that in the special case where t, S is the region under the graph of f and our general definition of area () reduces to our previous definition (Definition in Section 5.). 45

3 46 CHAPTER 6 APPLICATIONS OF INTEGRATION a b FIGURE 3 A=jj b ƒ d-j d a j b a T B Î T - B a b FIGURE 5 S = =ƒ FIGURE 4 FIGURE 6 = = Î (, ) Î T =- (, ) B = = In the case where both f and tare positive, ou can see from Figure 3 wh () is true: A area under f area under t b f d a b t d a b f t d a EXAMPLE Find the area of the region bounded above b e, bounded below b, and bounded on the sides b and. SOLUTION The region is shown in Figure 4. The upper boundar curve is e and the lower boundar curve is. So we use the area formula () with f e,, a, and b : t A e d e ] e e.5 M In Figure 4 we drew a tpical approimating rectangle with width as a reminder of the procedure b which the area is defined in (). In general, when we set up an integral for an area, it s helpful to sketch the region to identif the top curve T, the bottom curve B, and a tpical approimating rectangle as in Figure 5. Then the area of a tpical rectangle is T B and the equation A nl lim n T B b i a T B d summarizes the procedure of adding (in a limiting sense) the areas of all the tpical rectangles. Notice that in Figure 5 the left-hand boundar reduces to a point, whereas in Figure 3 the right-hand boundar reduces to a point. In the net eample both of the side boundaries reduce to a point, so the first step is to find a and b. V EXAMPLE Find the area of the region enclosed b the parabolas and. SOLUTION We first find the points of intersection of the parabolas b solving their equations simultaneousl. This gives, or. Thus, so or. The points of intersection are, and,. We see from Figure 6 that the top and bottom boundaries are T and B The area of a tpical rectangle is T B and the region lies between and. So the total area is A d d 3 3 M

4 .5 _ FIGURE (mi/h) FIGURE 8 = _ A œ + B =$ t (seconds) Sometimes it s difficult, or even impossible, to find the points of intersection of two curves eactl. As shown in the following eample, we can use a graphing calculator or computer to find approimate values for the intersection points and then proceed as before. EXAMPLE 3 Find the approimate area of the region bounded b the curves s and 4. SOLUTION If we were to tr to find the eact intersection points, we would have to solve the equation s 4 This looks like ver difficult equation to solve eactl (in fact, it s impossible), so instead we use a graphing device to draw the graphs of the two curves in Figure 7. One intersection point is the origin. We zoom in toward the other point of intersection and find that.8. (If greater accurac is required, we could use Newton s method or a rootfinder, if available on our graphing device.) Thus an approimation to the area between the curves is.8 A s 4 d To integrate the first term we use the subsitution u. Then du d, and when.8, we have u.39. So A.39 du su.8 4 d su ].39 5 SECTION 6. AREAS BETWEEN CURVES s EXAMPLE 4 Figure 8 shows velocit curves for two cars,a and B, that start side b side and move along the same road. What does the area between the curves represent? Use the Midpoint Rule to estimate it. SOLUTION We know from Section 5.4 that the area under the velocit curve A represents the distance traveled b car A during the first 6 seconds. Similarl, the area under curve B is the distance traveled b car B during that time period. So the area between these curves, which is the difference of the areas under the curves, is the distance between the cars after 6 seconds. We read the velocities from the graph and convert them to feet per second mi h ft s. t va vb va vb M

5 48 CHAPTER 6 APPLICATIONS OF INTEGRATION S a b FIGURE 9 = S S =ƒ We use the Midpoint Rule with n 4 intervals, so that t 4. The midpoints of the intervals are t, t 6, t3, and t4 4. We estimate the distance between the cars after 6 seconds as follows: 6 va vb dt t ft M If we are asked to find the area between the curves f and t where f t for some values of but t f for other values of, then we split the given region S into several regions S, S,... with areas A, A,... as shown in Figure 9. We then define the area of the region S to be the sum of the areas of the smaller regions S, S,..., that is, A A A. Since f t f t when f t t f when t f we have the following epression for A. 3 The area between the curves f and t and between a and b is A b a f t d FIGURE =cos =sin A A = = π π π 4 When evaluating the integral in (3), however, we must still split it into integrals corresponding to A, A,.... V EXAMPLE 5 Find the area of the region bounded b the curves sin, cos,, and. SOLUTION The points of intersection occur when sin cos, that is, when 4 (since ). The region is sketched in Figure. Observe that cos sin when 4 but sin cos when 4. Therefore the required area is A cos sin d A A 4 cos sin d sin cos d 4 4 [sin cos ] [ cos sin ] 4 s s s s s In this particular eample we could have saved some work b noticing that the region is smmetric about 4 and so 4 cos sin d A A M

6 Some regions are best treated b regarding as a function of. If a region is bounded b curves with equations f, t, c, and d, where f and tare continuous and f t for c d (see Figure ), then its area is d c =g() =d =c Î =f() A d c SECTION 6. AREAS BETWEEN CURVES 49 f t d d L R - L R Î c 4 L = -3 (_, _) FIGURE 3 = œ +6 3 A =_ œ +6 FIGURE 4 _ A (5, 4) R =+ (_, _) =- (5,4) FIGURE If we write R for the right boundar and L for the left boundar, then, as Figure illustrates, we have A d c R L d Here a tpical approimating rectangle has dimensions R L and. V EXAMPLE 6 Find the area enclosed b the line and the parabola 6. SOLUTION B solving the two equations we find that the points of intersection are, and 5, 4. We solve the equation of the parabola for and notice from Figure 3 that the left and right boundar curves are L 3 R We must integrate between the appropriate -values, and 4. Thus A 4 R L d 4 [ ( 3)]d 4 ( 4)d 3 3 FIGURE ( 4 3 8) 8 We could have found the area in Eample 6 b integrating with respect to instead of, but the calculation is much more involved. It would have meant splitting the region in two and computing the areas labeled A and A in Figure 4. The method we used in Eample 6 is much easier. 4 M

7 4 CHAPTER 6 APPLICATIONS OF INTEGRATION 6. 4 Find the area of the shaded region... =5-3. EXERCISES = - =_ = (4,4) = =e 5 8 Sketch the region enclosed b the given curves. Decide whether to integrate with respect to or. Draw a tpical approimating rectangle and label its height and width. Then find the area of the region. 5., 9,, 6. sin, e,, 7., 8., 4 9.,,. s, 3 3.,., 4 3., 6 4. cos, cos, 5. tan, sin, , 3 7. s,, ,, 3, 3 9., 4. 4, 4. (_3,3) = -4 =- =œ + = + =.,. sin, 3. cos, sin,, 4. cos, cos, 5., 6., 7.,, 4, 8. 3, 8, 4 4, 9 3 Use calculus to find the area of the triangle with the given vertices. 9.,,,,, 6 3., 5,,, 5, 3 3 Evaluate the integral and interpret it as the area of a region. Sketch the region. 3. sin cos d 3. 4 s d Use the Midpoint Rule with n 4 to approimate the area of the region bounded b the given curves. 33. sin 4, cos 4, 34. s 3 6 3,, ; Use a graph to find approimate -coordinates of the points of intersection of the given curves. Then find (approimatel) the area of the region bounded b the curves. 35. sin, e, 37. 3, cos,

8 CAS 39. Use a computer algebra sstem to find the eact area enclosed b the curves and. 4. Sketch the region in the -plane defined b the inequalities, and find its area. 4. Racing cars driven b Chris and Kell are side b side at the start of a race. The table shows the velocities of each car (in miles per hour) during the first ten seconds of the race. Use the Midpoint Rule to estimate how much farther Kell travels than Chris does during the first ten seconds. t vc vk t vc vk The widths (in meters) of a kidne-shaped swimming pool were measured at -meter intervals as indicated in the figure. Use the Midpoint Rule to estimate the area of the pool A cross-section of an airplane wing is shown. Measurements of the height of the wing, in centimeters, at -centimeter intervals are 5.8,.3, 6.7, 9., 7.6, 7.3, 3.8,.5, 5., 8.7, and.8. Use the Midpoint Rule to estimate the area of the wing s cross-section. cm 44. If the birth rate of a population is b t e people per ear and the death rate is d t 46e people.4t per ear, find the area between these curves for.8t t. What does this area represent? 45. Two cars,a and B, start side b side and accelerate from rest. The figure shows the graphs of their velocit functions. (a) Which car is ahead after one minute? Eplain. (b) What is the meaning of the area of the shaded region? SECTION 6. AREAS BETWEEN CURVES 4 (c) Which car is ahead after two minutes? Eplain. (d) Estimate the time at which the cars are again side b side. A 46. The figure shows graphs of the marginal revenue function R and the marginal cost function C for a manufacturer. [Recall from Section 4.7 that R and C represent the revenue and cost when units are manufactured. Assume that R and C are measured in thousands of dollars.] What is the meaning of the area of the shaded region? Use the Midpoint Rule to estimate the value of this quantit. 3 B Rª() Cª() 5 ; 47. The curve with equation is called Tschirnhausen s cubic. If ou graph this 3 curve ou will see that part of the curve forms a loop. Find the area enclosed b the loop. 48. Find the area of the region bounded b the parabola, the tangent line to this parabola at,, and the -ais. 49. Find the number b such that the line divides the region bounded b the curves and into two regions with equal area. b 4 5. (a) Find the number a such that the line bisects the area under the curve, (b) Find the number b such that the a 4. line b bisects the area in part (a). 5. Find the values of c such that the area of the region bounded b the parabolas c and c is Suppose that c. For what value of c is the area of the region enclosed b the curves cos, cos c, and equal to the area of the region enclosed b the curves cos c,, and? 53. For what values of m do the line and the curve m enclose a region? Find the area of the region. t(min)