APPM 1350 Review Solutions 3

# 52 u 5 1 2 0 2 7 2 5 8 35 h if we use the

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Unformatted text preview: ge of variables u = x2 , then du = 2xdx and so we have Z1 p Z 1p 4 dx = 2 4x 1 x 1 u2 du Z 0 1p 0 now note that the integral 1 u2 du represents the area of one quarter of a unit circle, since 0 p the graph of y = 1 u2 is the top half of a circle with radius equal to one. Thus, using the formula A = ⇡ r2 we have, Z1 p Z 1p ⇡ ⇡ 4x 1 x4 dx = 2 1 u2 du = 2 · = 4 2 0 0 x2 Z 2. (a) If f is continuous and 4 f (x) dx = 10, ﬁnd 0 (b) If f 0 is continuous on [a, b], show that 2 Z b Z 2 f (2x) dx. 0 f (x)f 0 (x)dx = [f (b)]2 [f (a)]2 . a Solution: (a) Let u = 2x, then du = 2dx and so Z2 Z 14 f (2x) dx = f (u) du = 5. 20 0 (b) Note that since f 0 is continuous that implies the product f (x)f 0 (x) is continuous and therefore it is integrable, so the integral makes sense. Now if we let u = f (x) then du = f 0 (x)dx and so Zb Z f (b ) u 2 f (b ) 2 f (x)f 0 (x)dx = 2 u du = 2 · = [f (b)]2 [f (a)]2 2 f (a ) a f (a) p 3. (a) Find the average value of f (t) = t 1 + t2 over the interval [0,...
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## This note was uploaded on 02/04/2014 for the course APPM 1350 taught by Professor Segur,harv during the Fall '07 term at Colorado.

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