# Not too sure what to do

• Jun 1st 2006, 05:28 PM
Nichelle14
Not too sure what to do
Compute integral of x*f'(x)*f(x) dx from a to b if

the integral of f^2(x)dx from a to b equals 2 and f(a)=f(b)=0. :confused:
• Jun 1st 2006, 08:20 PM
malaygoel
Quote:

Originally Posted by Nichelle14
Compute integral of x*f'(x)*f(x) dx from a to b if

the integral of f^2(x)dx from a to b equals 2 and f(a)=f(b)=0. :confused:

use integration by parts
integral of {x}*{f'(x)*f(x)} dx from a to b
={x*f^2(x)/2 from a to b} - {integral of 1*f^2(x)/2 dx from a to b}
={b*f^2(b)- a*f^2(a)} -2/2
=b*0 -a*0 -1
= -1
• Jun 1st 2006, 08:54 PM
Soroban
Hello, Nichelle14!

Quote:

Compute: $I \;= \;\int^b_a x\cdot f'(x)\cdot f(x)\,dx$

if $\int^b_a [f(x)]^2 = 2$ and $f(a) = f(b) = 0$
Integrate by parts . . .

Let: $u = x\qquad\qquad dv = f(x)\cdot f'(x)\,dx$

Then: $du = dx\qquad v = \frac{1}{2}[f(x)]^2$

We have: . $I \;= \;\underbrace{\frac{1}{2}x\left[f(x)\right]^2\bigg|^b_a} - \frac{1}{2}\underbrace{\int^b_a[f(x)]^2\,dx}$

. . $I \;= \;\overbrace{\frac{1}{2}(b)[f(b)]^2 - \frac{1}{2}(a)[f(a)]^2} - \frac{1}{2}[2]$

. . $I \;= \;\frac{1}{2}(b)(0^2) - \frac{1}{2}(a)(0^2) - 1 \;\;= \;0 - 0 -1 \;\;=$ -1