# Integration By Parts

1. Feb 9, 2006

### Mindscrape

Okay, so here is the problem I have, which I am getting tripped up on for some reason:
a) Use integration by parts to show that
$$\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{a}^{b} xf'(x) dx$$
this was pretty easy, just regular old integration by parts with limits of integration.

b) Use the result in part (a) to show that if y = f(x) then
$$\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(y) dy$$
I know that the inverse function of f(x) will leave me with x, which gets me to
$$\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{a}^{b} f^{-1}(y) f'(x)$$
but then what happens with f'(x)? For some reason I'm just totally spacing this out.

2. Feb 9, 2006

### StatusX

You want the f'(x). It appears in your first formula. Just simplify f-1(y).

3. Feb 10, 2006

### Mindscrape

I think you are working the opposite way. Basically what I'm trying to do is get from
$$\int_a^b f(x)dx$$

to (if we let $$y = f(x)$$)

$$\int_{a}^{b} f(x) dx = bf(b) - af(a) - \int_{f(a)}^{f(b)} f^{-1}(y) dy$$

to (if we let $$\alpha = f(a)$$ and $$\beta = f(b)$$)

$$\int_{\alpha}^{\beta} f^{-1}(x)dx = \beta f^{-1}(\beta) - \alpha f^{-1}(\alpha) - \int_{f^{-1}(\alpha)}^{f^{-1}(\beta)} f(x)dx$$

Last edited: Feb 10, 2006
4. Feb 10, 2006

### NateTG

From part a it should be obvious that it's sufficient to show that the two integrals are equal.

So, what happens if you try to integrate
$$\int_a^b xf'(x) dx$$
by subsitution using
$$u=f(x)$$
?

Last edited: Feb 10, 2006