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Unformatted text preview: k =1 (right) ∆x = n n k =1 lim n→+∞ k =1 192 384(2) 256 + + = 320 2 6 4 2k −3 n 1− k =1 n 3 56 2 = n n n 1− k =1 108 n2 n k+ k =1 108 n(n + 1) 72 n(n + 1)(2n + 1) 16 +3 −4 = 56 − 2 n 2 n 6 n 72(2) 16 ∗ − = 22 f (xk )∆x = 56 − 54 + 6 4 (left) ∆x = n f (x∗ k )∆x = k =1 2(k − 1) −3 n 1− k =1 = 56 n n 1− k =1 108 n2 3 n→+∞ k =1 n k2 − k =1 n(n + 1) 2 n 16 n4 k3 k =1 2 ; n (k − 1) + k =1 2 n 72 n3 n (k − 1)2 − k =1 16 n4 108 (n − 1)n 72 (n − 1)n(2n − 1) 16 +3 −4 n2 2 n 6 n 108 72(2) 16 + − = 22 f (x∗ )∆x = 56 − k 2 6 4 = 56 − n 72 n3 b−a 2(k − 1) 2 = , x∗ = −3 + , n nk n n lim (n − 1)n 2 2 2k b−a = , x∗ = −3 + , k n n n f (x∗ )∆x = k (b) ; 192 384(2) 256 + + = 320 2 6 4 b−a 4 4(k − 1) = , x∗ = 2 + , n nk n n 3 192 (n − 1)n 384 (n − 1)n(2n − 1) 256 4(k − 1) 4 = 32+ 2 +3 +4 2+ f (x∗ )∆x = k n n n 2 n 6 n lim (a) 2 (left) ∆x = n 44. n(n + 1) 2 n (k − 1)3 k =1 (n − 1)n 2 2 ; (a) f is continuous on [−1, 1] so f is integrable there by part (a) of Theorem 7.5.8 (b) |f (x)| ≤ 1 so f is bounded on [−1, 1], and f has one point of discontinuity, so by part (b) of Theorem 7.5.8 f is integrable on [−1, 1] (c) 45. f is not bounded on [-1,1] because lim f (x) = +∞, so f is not integrable on [0,1] (d) 46. x→0 1 does not exist. f is continuous x elsewhere. −1 ≤ f (x) ≤ 1 for x in [−1, 1] so f is bounded there. By part (b), Theorem 7.5.8, f is integrable on [−1, 1]. f (x) is discontinuous at the point x = 0 because lim sin x→0 Each subinterval of a partition of [a, b] contains both rational and irrational numbers. If all x∗ are k chosen to be rational then n n f (x∗ )∆xk k k =1 = n n ∆xk = b − a so (1)∆xk = k =1 k =1 f (x∗ )∆xk = b − a. k lim max ∆xk →0 k =1 n If all x∗ are irrational then k preceding limits are not equal. f (x∗ )∆xk = 0. f is not integrable on [a, b] because the k lim max ∆xk →0 k =1 2 ; 229 Chapter 7 n 47. (a) lim max ∆xk →0 cf (x∗ )∆xk = cSn and we want to prove k f (x)dx then a k =1 that n b f (x∗ )∆xk and S = k Let Sn = k =1 cSn = cS . If c = 0 the result follows immediately, so suppose that c = 0 then for any > 0, |cSn − cS | = |c||Sn − S | < if |Sn − S | < /|c|. But because f is integrable on [a, b], there is a number δ > 0 such that |Sn − S | < /|c| whenever max ∆xk < δ so |cSn − cS | < and hence lim cSn = cS . max ∆xk →0 n (b) n f (x∗ )∆xk , Sn = k Let Rn = k =1 n g (x∗ )∆xk , Tn = k k =1 [f (x∗ ) + g (x∗ )]∆xk , R = k k k =1 b f (x)dx, and a b g (x)dx then Tn = Rn + Sn and we want to prove that S= a lim max ∆xk →0 Tn = R + S . |Tn − (R + S )| = |(Rn − R) + (Sn − S )| ≤ |Rn − R| + |Sn − S | so for any > 0 |Tn − (R + S )| < if |Rn − R| + |Sn − S | < . Because f and g are integrable on [a, b], there are numbers δ1 and δ2 such that |Rn − R| < /2 for max ∆xk < δ1 and |Sn − S | < /2 for max ∆xk < δ2 . If δ = min(δ1 , δ2 ) then |Rn − R| < /2 and |Sn − S | < /2 for max ∆xk < δ thus |Rn − R| + |Sn − S | < and so |Tn − (R + S )| < for max ∆xk < δ which shows that lim Tn = R + S . max ∆xk →0 48. For the smallest, find x∗ so that f (x∗ ) is minimum on each subinterval: x∗ = 1, x∗ = 3/2, x∗ = 3 so 1 2 3 k k (2)(1) + (7/4)(2) + (4)(1) = 9.5. For the largest, find x∗ so that f (x∗ ) is maximum on each subinterval: k k x∗ = 0, x∗ = 3, x∗ = 4 so (4)(1) + (4)(2) + (8)(1) = 20. 1 2 3 EXERCISE SET 7.6 2 1. (a) (2 − x)dx = (2x − x2 /2) 0 −1 2dx = 2x −1 = 2(1) − 2(−1) = 4 3 (x + 1)dx = (x2 /2 + x) (c) 1 2 xdx = x /2 (a) 3 1 −1 = 81/4 − 16/4 = 65/4 2 1 2 xdx = x3/2 3 9. 2 = (27 − 1) = 52/3 3 1 −1 4 6. x =e −e x 3 e dx = e 1 5 8. 1 1 13 x − 2x2 + 7x 3 0 = 48 −3 10. x4 dx = x5 /5 −1 −3/5 x 3 3 1 9 1 1 4. 2 5. 7. = 4/2 + 6 − (1/2 − 3) = 21/2 3 x3 dx = x4 /4 √ 3 2 (x + 3)dx = (x2 /2 + 3x) 3 9 = 5(9) − 5(3) = 30 5dx = 5x 3 2 3. (b) 0 −1 9 9 = 25/2 0 (c) = 9/2 + 3 − (1/2 + 1) = 6 5 5 2. = 4 − 4/2 = 2 1 1 (b) 2 0 5 dx = x2/5 2 1 dx = ln x x 12 15 x+ x 2 5 = 1/5 − (−1)/5 = 2/5 4 = 1 5 2/5 (4 − 1) 2 5 = ln 5 − ln 1 = ln 5 1 2 = 81/10 −1 Exercise Set 7.6 3 1 sin x 3 ln 2 16. √ 2 18. = √ 10 2 6 t − t3/2 + √ 3 t 26. 3/2 2 4a a = ln 2 + 1 3π/4 0 0 (b) −1 0 1 30. 4 1 3 31. 0.665867079; 1 0 √ 2/2 π/2 0 0 13 x 3 1 1 dx = − 2 x x π /2 32. 1.000257067; =2− 2 2 2 2 + x dx = − (2 − x)3/2 + (2 + x)3/2 3 3 −1 0 √ √ √ √ √ 2 2 2 = − (2 2 − 3 3) + (8 − 2 2) = (8 − 4 2 + 3 3) 3 3 3 0 −2 2 1 dx = x3/2 2 x 3 x dx + 3π/4 − sin x 0 √ = 9/4 + 1/4 = 5/2 3/2 π /2 (− cos x)dx = sin x (ex − 1)dx = (x − ex ) (−x)dx = 0 0 = 10819/324 4 2 + (x2 − 3x) 0 0 x2 dx + √ 1 (1 − ex )dx + 3 −2 2 2 − x dx + 0 29. (2x − 3)dx = (3x − x2 ) π/2 −1 9 √2 2(e − e) + csc 2 − csc 1 3/2 2 (3 − 2x)dx + √ = (ln 2)/2 4 2 √ 8 y + y 3/2 − 3/2 3 3y...
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