# 51 0 i a a a f a u du f a u f u du

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

Ask a homework question - tutors are online