How Does Integration by Parts Prove This Real Analysis Identity?

[steelphantom]

Homework Statement

Let f be of class C¹ on [a, b], with f(a) = f(b) = 0. Show that \int_a^b xf(x)f'(x)dx = -1/2 \int_a^b [f(x)]^2 dx.

Homework Equations

If F is an antiderivative of f, then \int_a^b f(t)dt = F(b) - F(a)

The Attempt at a Solution

I'm just really not sure how to begin this one. I know that because f is of class C¹ that f' is continuous. Maybe change of variables?

 

[sutupidmath]

Use integration by parts!

 

[Hurkyl]

Less dogmatically... one of the major differences between the two things you're trying to show equal is that one has a derivative in it, and the other doesn't. And what methods do you know that can increase/decrease how derivated a part of your integrand is?

 

[sutupidmath]

\int_a^b xf(x)f'(x)dx

u=x, du=dx, v=\int f(x)f'(x)dx

v=\int f(x)f'(x)dx, t=f(x), dt=f'(x)dx=>\int tdt=\frac{1}{2}t^2=\frac{1}{2}[f(x)]^2

\frac{1}{2}x[f(x)]|_a^b-\frac{1}{2}\int_a^b [f(x)]^2dx=-\frac{1}{2}\int_a^b [f(x)]^2dx

Because:\frac{1}{2}x[f(x)]|_a^b=0