# Real Analysis: Integration

1. Oct 29, 2008

### steelphantom

1. The problem statement, all variables and given/known data
Let f be of class C1 on [a, b], with f(a) = f(b) = 0. Show that $$\int_a^b xf(x)f'(x)dx$$ = $$-1/2 \int_a^b [f(x)]^2 dx$$.

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

3. 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?

2. Oct 29, 2008

### sutupidmath

Use integration by parts!

3. Oct 29, 2008

### Hurkyl

Staff Emeritus
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?

4. Oct 29, 2008

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