This preview shows page 1. Sign up to view the full content.
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.
 Fall '07
 SEGUR,HARV

Click to edit the document details