# Integration by parts / proof

1. May 7, 2005

If $$f(0)=g(0)=0$$, show that

$$\int _0 ^a f(x) g ^{\prime \prime} (x) \: dx = f(a) g^{\prime} (a) - f^{\prime} (a) g (a) + \int _0 ^a f ^{\prime \prime} (x) g (x) \: dx$$

I know I need to use integration by parts, but I'm having a hard time figuring out the right choice of $$u$$ and $$dv$$. What I do know is the following:

$$\underbrace{\int _0 ^a f(x) g ^{\prime \prime} (x) \: dx}_{\int _0 ^a u \: dv} = \underbrace{f(a) g^{\prime} (a) - f^{\prime} (a) g (a)}_{\left. uv \right] _0 ^a} + \underbrace{\int _0 ^a f ^{\prime \prime} (x) g (x) \: dx}_{-\int _0 ^a v \: du}$$

Also, the closest I get to $$\left. uv \right] _0 ^a$$ so far is

$$\frac{d}{dx} \left[ f(x) g(x) \right] = f(x) g^{\prime} (x) + f^{\prime} (x) g (x) \Longrightarrow \left. \frac{d}{dx} \left[ f(x) g(x) \right] \right|_0 ^a = f(a) g^{\prime} (a) + f^{\prime} (a) g (a)$$

but, as you can see, that doesn't work because of a sign.

Any help is highly appreciated.

2. May 7, 2005

### jdavel

thiago j,

Since you have g'' and f on the right, and g and f'' on the left, you have to integrate by parts twice. The first time your integrand you end up with on the right will be f'g'. Then integrate again and it will become f''g. Meanwhile you will have created two of the uv terms in front. If you're careful with the signs they'll work out just like the answer that's given.

One more hint. Do it with indefinite interals first to reduce the amount of stuff you have to keep track of at a time.

3. May 7, 2005

Thank you for the tips!!!

$$\int _0 ^a f(x) g^{\prime \prime} (x) \: dx$$

Integrating by parts gives

$$u = f(x) \Rightarrow \frac{du}{dx} = f ^{\prime} (x) \Rightarrow du = f ^{\prime} (x) \: dx$$
$$dv = g^{\prime \prime} (x) \: dx \Rightarrow v = g^{\prime} (x)$$

$$\int _0 ^a f(x) g^{\prime \prime} (x) \: dx = \left. f(x) g^{\prime} (x) \right] _0 ^a - \int _0 ^a f^{\prime} (x) g^{\prime} (x) \: dx$$
$$\int _0 ^a f(x) g^{\prime \prime} (x) \: dx = f(a) g^{\prime} (a) - \int _0 ^a f^{\prime} (x) g^{\prime} (x) \: dx$$

which implies we need to find

$$\int _0 ^a f^{\prime} (x) g^{\prime} (x) \: dx$$

Integrating by parts gives

$$w = f ^{\prime} (x) \Rightarrow \frac{dw}{dx} = f ^{\prime \prime} (x) \Rightarrow dw = f ^{\prime \prime} (x) \: dx$$
$$dk= g^{\prime} (x) \: dx \Rightarrow k = g(x)$$

$$\int _0 ^a f^{\prime} (x) g^{\prime} (x) \: dx = \left. f^{\prime} (x) g (x) \right] _0 ^a - \int _0 ^a f^{\prime \prime} (x) g(x) \: dx$$
$$\int _0 ^a f^{\prime} (x) g^{\prime} (x) \: dx = f^{\prime} (a) g (a) - \int _0 ^a f^{\prime \prime} (x) g(x) \: dx$$

Thus, we obtain

$$\int _0 ^a f(x) g^{\prime \prime} (x) \: dx & = f(a) g^{\prime} (a) - f^{\prime} (a) g (a) + \int _0 ^a f^{\prime \prime} (x) g(x) \: dx$$