Y f x x3 for 1 x 2 f x 3x2 2 1 9x4 32 2x3

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Unformatted text preview: x > 0. from part (b), e1/(x+1) < eln(1+1/x) < e1/x , e1/(x+1) < 1 + 1/x < e1/x , ex/(x+1) < (1 + 1/x)x < e; by the Squeezing Theorem, lim (1 + 1/x)x = e. x→+∞ (d) 45. Use the inequality ex/(x+1) < (1 + 1/x)x to get e < (1 + 1/x)x+1 so (1 + 1/x)x < e < (1 + 1/x)x+1 . From Exercise 44(d) e − 1 + 1 50 50 < y (50), and from the graph y (50) < 0.06 0.2 0 100 0 46. F (x) = f (x), thus F (x) has a value at each x in I because f is continuous on I so F is continuous on I because a function that is differentiable at a point is also continuous at that point CHAPTER 7 SUPPLEMENTARY EXERCISES 5. If the acceleration a = const, then v (t) = at + v0 , s(t) = 1 at2 + v0 t + s0 2 6. (a) Divide the base into n equal subintervals. Above each subinterval choose the lowest and highest points on the curved top. Draw a rectangle above the subinterval going through the lowest point, and another through the highest point. Add the rectangles that go through the lowest points to obtain a lower estimate of the area; add the rectangles through the highest points to obtain an upper estimate of the area. Supplementary Exercises 7 250 (b) (c) (a) 3 11 += 24 4 5 −1 − (e) 8. n = 20: 24.4 cm, 23.1 cm (c) 7. n = 10: 25.0 cm, 22.4 cm not enough information (a) (c) 3 4 =− (a) (b) (c) 10. 35 4 −1 1 dx + (f ) not enough information (b) not enough information 13 1 (d) 4(2) − 3 = 2 2 3 − π (3)2 /4 = 0 u = x2 , du = 2xdx; 1 2 −2 1 − x2 dx = 2(1) + π (1)2 /2 = 2 + π/2 −1 12 (x + 1)3/2 3 1 3 =− 2 2 (d) 5 1 +2= 2 2 not enough information 1 9. (b) −1 − 1 1 2 1 (103/2 − 1) − 9π/4 3 1 − u2 du = 0 1 π (1)2 /4 = π/8 2 y 1 0.8 0.6 0.4 0.2 x 0.2 0.6 1 11. The rectangle with vertices (0, 0), (π, 0), (π, 1) and (0, 1) has area π and is much too large; so is the triangle with vertices (0, 0), (π, 0) and (π, 1) which has area π/2; 1 − π is negative; so the answer is 35π/128. 12. Divide ex + 3 into e2x to get e2x dx = ex + 3 13. ex dx − 3 e2x 3ex = ex − x so ex + 3 e +3 ex dx = ex − 3 ln(ex + 3) + C ex + 3 Since y = ex and y = ln x are inverse functions, their graphs are symmetric with respect to the line y = x; consequently the areas A1 and A3 are equal (see figure). But A1 + A2 = e, so e y e A1 1 x ln xdx + 1 e dx = A2 + A3 = A2 + A1 = e 1 0 A2 A3 1 14. (a) 1 n n k =1 1 lim n→+∞ n k = n n k =1 n f (x∗ )∆x where f (x) = k k =1 k = n 1 x1/2 dx = 0 2 3 √ e x x, x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus k 251 Chapter 7 (b) 1 n n k =1 1 lim n→+∞ n n (c) k =1 n 15. k =1 1 x4 dx = = 0 1 5 n f (x∗ )∆x where f (x) = ex , x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus k k k =1 k/n e lim k =1 4 k n k =1 n f (x∗ )∆x where f (x) = x4 , x∗ = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus k k = ek/n = n n→+∞ n 4 k n n n→+∞ n 1 f (x∗ )∆x = k = lim ex dx = e − 1. 0 k =1 1 is positive and increasing on the interval [1, 2], the left endpoint approximation x 1 overestimates the integral of and the right endpoint approximation underestimates it. x Since f (x) = (a) For n = 5 this becomes 1 1 1 1 1 + + + + < 0.2 1.2 1.4 1.6 1.8 2.0 2 1 1 1 1 1 1 1 dx < 0.2 + + + + x 1.0 1.2 1.4 1.6 1.8 2 (b) 1 dx = ln 2 is x For general n the left endpoint approximation to 1 1 n n k =1 n k =1 n 1 = 1 + (k − 1)/n k =1 n 1 . This yields n+k k =1 1 = n+k−1 1 < n+k 2 1 n−1 k =0 1 and the right endpoint approximation is n+k n−1 1 dx < x k =0 1 which is the desired inequality. n+k 1 1 1 1 = so ≤ 0.1, n ≥ 5 (c) By telescoping, the difference is − n 2n 2n 2n (d) n ≥ 1, 000 16. The direction field is clearly an even function, which means that the solution is even, its derivative is odd. Since sin x is periodic and the direction field is not, that eliminates all but x, the solution of which is the family y = x2 /2 + C . 17. (a) n 1 · 2 + 2 · 3 + · · · + n(n + 1) = n k =1 = n−1 (b) k =1 k 9 − n n2 17 lim n→+∞ 2 3 n−1 n 2 (c) = j =1 n−1 1− k =1 3 j= j =1 1 n2 (a) n−1 k= k =1 11 17 9 (n − 1) − 2 · (n − 1)(n) = n n2 2 i=1 1 2i + (2)(3) = 2 2 3 3 i+ n−1 ; n i=1 i=1 1 3 = 2 · (3)(4) + (3)(3) = 21 2 19 (k + 4)(k + 1) k =0 k =1 1 1 1 n(n + 1)(2n + 1) + n(n + 1) = n(n + 1)(n + 2) 6 2 3 14 18. k k =1 17 = 2 2 i+ i=1 9 n n k2 + k (k + 1) = (k − 1)(k − 4) (b) k =5 Supplementary Exercises 7 19. (a) 252 If u = sec x, du = sec x tan xdx, sec2 x tan xdx = if u = tan x, du = sec2 xdx, (b) 20. 1 2 π /4 = 0 udu = u2 /2 + C2 = (tan2 x)/2 + C2 . x−1/3 1 2 tan2 x π /4 0 = 1 (1 − 0) = 1/2 2 x2/3 + 1dx; u = x2/3 + 1, du = 2 −1/3 x dx 3 u1/2 du = u3/2 + C = (x2/3 + 1)3/2 + C n n b 22. 1 (2 − 1) = 1/2 and 2 1 + x−2/3 dx = 3 2 udu = u2 /2 + C1 = (sec2 x)/2 + C1 ; They are equal only if sec2 x and tan2 x differ by a constant, which is true. sec2 x 21. sec2 x tan xdx = b fk (x)dx = (a) a k =1 fk (x)dx k =1 a b (b) yes; substitute ck fk (x) for fk (x) in part (a), and then use a Theorem 7.5.4 x 23. (a) 1 fk (x)dx from a 1 d...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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