APPM 1350 Review Solutions 3

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

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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,...
View Full Document

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.

Ask a homework question - tutors are online