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HW3-510-soln - AMS 510 HW3 Solution 5.34 1 lim LHS = n n...

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AMS 510 HW3 Solution 5.34 LHS = lim n ∞ 1 n i = 1 n 1 1 i 2 n 2 = 0 1 1 1 x 2 dx = 4 5.35 LHS = lim n ∞ 1 n i = 1 n i n p = 0 1 x p dx = 1 p 1 5.38 LHS = lim n ∞ 1 n i = 1 n 1 1  i n 2 = 0 1 1 1  x 2 dx = ln 1 2 5.44 0 1 cos nx x 1 dx 0 1 cos nx x 1 dx 0 1 1 x 1 dx as 0 x 1 , LHS = 0 1 1 x 1 dx = ln 2 , so we get the conclusion. 5.47 when 0 x 1 , both sin x and 1 x 2 1 don't change sign, so from generalized mean value theorem, 0 1 sin x x 2 1 d x = 0 1 sin x 1 x 2 1 d x = sin  2 0 1 1 x 2 1 d x = 4 sin  2 0 1 sin x x 2 1 d x = 0 1 sin x 1 x 2 1 d x = 1 1 2 1 0 1 sin x d x = 2  1 2 1 5.55 (a)
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0 1 x 2 tan 1 x dx = 0 1 1 3 tan 1 x dx 3 = 1 3 x 3 tan 1 x 0 1 0 1 1 3 x 3 1 x 2 dx = 12 0 1 1 3 x x 1 x 2 dx = 12 1 6 0 1 1 3 x 1 x 2 dx = 12 1 6 1 6 ln 2 (b) 2 2 x 2 x 1 dx = 2 2 x 1 2 2 3 4 dx = 2 2 x 1 2 2  3 2 2 dx , from NO. 31 integral formula, we can
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