FDWK_3ed_Ch05_pp262-319

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Section 5.4 Fundamental Theorem of Calculus 301 How to Find Total Area Analytically To find the area between the graph of y f x and the x-axis over the interval a, b analytically, 1. partition a, b with the zeros of f, 2. integrate f over each subinterval, 3. add the absolute values of the integrals. We can find area numerically by using NINT to integrate the absolute value of the function over the given interval. There is no need to partition. By taking absolute values, we automatically reflect the negative portions of the graph across the x-axis to count all area as positive (Figure 5.29). [–3, 3] by [–3, 3] (a) EXAMPLE 7 Finding Area Using NINT Find the area of the region between the curve y x cos 2x and the x-axis over the interval 3 x 3 (Figure 5.29). SOLUTION Rounded to two decimal places, we have NINT x cos 2x, x, 3, 3 5.43. Now try Exercise 51. [–3, 3] by [–3, 3] (b) Figure 5.29 The graphs of (a) y x cos 2x and (b) y x cos 2x over 3, 3. The shaded regions have the same area. y x 0 3 6 Figure 5.30 The graph of f in Example 8, in which questions are asked about the function h(x) x 1 f t dt. How to Find Total Area Numerically To find the area between the graph of y f x and the x-axis over the interval a, b numerically, evaluate NINT f x, x, a, b. Analyzing Antiderivatives Graphically A good way to put several calculus concepts together at this point is to start with the graph of a function f and consider a new function h defined as a definite integral of f. If h(x) x f t dt, for example, the Fundamental Theorem guarantees that h(x) f (x), so a the graph of f is also the graph of h. We can therefore make conclusions about the behavior of h by considering the graphical behavior of its derivative f, just as we did in Section 4.3. EXAMPLE 8 Using the Graph of f to Analyze h(x) x f(t) dt a The graph of a continuous function f with domain [0, 8] is shown in Figure 5.30. Let h be the function defined by h(x) x f t dt. 1 (a) Find h(1). (b) Is h(0) positive or negative Justify your answer. (c) Find the value of x for which h(x) is a maximum. (d) Find the value of x for which h(x) is a minimum. (e) Find the x-coordinates of all points of inflections of the graph of y h(x). continued
302 Chapter 5 The Definite Integral SOLUTION First, we note that h(x) f (x), so the graph of f is also the graph of the derivative of h. Also, h is continuous because it is differentiable. (a) h(1) 1 f t dt 0. 1 (b) h(0) 0 f t dt 0, because we are integrating from right to left under a positive 1 function. (c) The derivative of h is positive on (0,1), positive on (1, 4), and negative on (4, 8), so the continuous function h is increasing on [0, 4] and decreasing on [4, 8]. Thus f (4) is a maximum. (d) The sign analysis of the derivative above shows that the minimum value occurs at an endpoint of the interval [0, 8]. We see by comparing areas that h(0) 0 f t dt 0.5, 1 while h(8) = 8 f t dt is a negative number considerably less than 1. Thus f (8) is a 1 minimum. (e) The points of inflection occur where h f changes direction, that is, at x 1, x 3, and x 6. Now try Exercise 57. Quick Review 5.4 (For help, go to Sections 3.6, 3.7, and 3.9.) In Exercises 1–10, find dydx. 1. y sin x 2 2x cos x 2 2. y sin x 2 2 sin x cos x 3. y sec 2 x tan 2 x 0 4. y ln 3x ln 7x 0 5. y 2 x 2 x 1 ln 2 6. y x 2 x 7. y co s x x sin x cos x 8. y sin t and x cos t x x 2 cot t 9. xy x y 2 y 1 1 10. dxdy 3x 2 y x 3 x Section 5.4 Exercises In Exercises 1–20, find dy/dx. x 1. y 0 x 3. y 0 x 5. y 2 x 7. y 7 x 9. y 2 0 5x 11. y 1 u 2 2 u x (sin 2 t) dt sin 2 x 2. y (3t + cos t 2 ) dt 2 x (t 3 t) 5 dt (x 3 – x) 5 4. y 1 e 5 t dt 2 x (tan 3 u)du tan 3 x 6. y 4 x 1 t 1 x dt 8. y 1 t2 1 x2 p x e t2 dt 2xe x4 10. y 2 6 13. y ln (1 t 2 ) dt x 6 e u sec u du 2 sin t dt 3 cos t e x sec x cot 3t dt 2x cot 3x 2 3x cos x 2 1 e 5x 2 sin x 3 cos x 1 sin du 12. y px u du p 1 cos 2 u 1 x 25 2 x 7 14. y 2t 4 t 1 dt ln(1 + x 2 ) x 2x x 4 1 1 sin p x) 12. 1 cos (( 22 p x) 250x 5 cos t 15. y dt 16. y x 3 t 2 5 100x 3 90 25 t2 2t 9 125x 6 6 2 5x 2 t 3 dt 6 0 10 17. y 3x 2 cos x 3 x6 2 sin(r 2 ) dr 18. y ln (2 p 2 ) dp x 3x sin x 2 x 19. 3 6x ln(2 9x cos x cos(2t) dt 20. y 4 ) 2 x t 2 dt x 2 sin x 3x 2 cos(2x 3 ) 2x cos(2x 2 ) sin x cos In Exercises 21–26, construct a function of the form 2 x cos x sin 2 x y x f t dt C that satisfies the given conditions. a 21. d y sin 3 x, and y 0 when x 5. y x 5 dx sin3 t dt 22. d y e x tan x, and y 0 when x 8. dx y x 8 et tan t dt 23. d y ln(sin x 5), and y 3 when x 2. dx y x ln (sin t 5) dt 3 24. d 2 y 3 s co, x and y 4 when x 3. dx y x 3 s co t dt 4 3
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FDWK_3ed_Ch05_pp262-319

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