# Thread: integration by parts problem

1. ## integration by parts problem

Suppose that f(x) and g(x) are functions with f(0)=g(0)=0, f(1)=g(1)=0, and with continuous second derivatives.

Use integration by parts to show that

$

\int f''(x)g(x)\,dx = \int f(x)g''(x)\,dx
$

Then come up with specific examples for f(x) and g(x) that satisfy the above.

I tried using
u=g(x) v=f(x)
du=g'(x)dx dv=f''(x)dx

then

u=g'(x) v=f(x)
du=g''(x)dx dv=f'(x)dx

but I ended up with

$

\int f''(x)g(x)\,dx = g(x)f'(x)-g'(x)f(x)- \int f(x)g''(x)\,dx

$

and I don't know how to get rid of the two middle functions...

if I switched what U was for the second time I applied it, I get back to where I began and If I apply IBP again, i get into 3rd derivatives...

I'm really confused. Can someone give me some advice?

2. Originally Posted by valkyrie
Suppose that f(x) and g(x) are functions with f(0)=g(0)=0, f(1)=g(1)=0, and with continuous second derivatives.

Use integration by parts to show that

$

\int f''(x)g(x)\,dx = \int f(x)g''(x)\,dx
$

Then come up with specific examples for f(x) and g(x) that satisfy the above.

I tried using
u=g(x) v=f(x)
du=g'(x)dx dv=f''(x)dx

then

u=g'(x) v=f(x)
du=g''(x)dx dv=f'(x)dx

but I ended up with

$

\int f''(x)g(x)\,dx = g(x)f'(x)-g'(x)f(x)- \int f(x)g''(x)\,dx

$

and I don't know how to get rid of the two middle functions...

if I switched what U was for the second time I applied it, I get back to where I began and If I apply IBP again, i get into 3rd derivatives...

I'm really confused. Can someone give me some advice?
I am guessing that the integrals have limits of integration, and the they are from 0 to 1 or the statement is not true.

dont forget to evaluate the middle parts and the conditions that f and g are = 0 at 1 and 0 will make them go away.

3. Hmmm, thanks. they are from 0 to 1 but I couldnt get the BB code to work properly!

I don't really remember in such an abstract situation... how do I evaluate the middle parts? do I just set them from 0 to 1?

4. Originally Posted by valkyrie
Hmmm, thanks. they are from 0 to 1 but I couldnt get the BB code to work properly!

I don't really remember in such an abstract situation... how do I evaluate the middle parts? do I just set them from 0 to 1?
$g(x)f'(x)\bigg|_{0}^{1}-g'(x)f(x)\bigg|_{0}^{1}=g(1)f'(1)-g(0)f'(0)-(g'(1)f(1)-g'(0)f(0))$

but since $f(0)=f(1)=g(0)=g(1)=0$

$g(1)f'(1)-g(0)f'(0)-(g'(1)f(1)-g'(0)f(0))=0f'(1)-0f'(0)-g'(1)0+g'(0)0=0$

This is what we wanted yay

5. Originally Posted by TheEmptySet
$g(x)f'(x)\bigg|_{0}^{1}-g'(x)f(x)\bigg|_{0}^{1}=g(1)f'(1)-g(0)f'(0)-(g'(1)f(1)-g'(0)f(0))$

but since $f(0)=f(1)=g(0)=g(1)=0$

$g(1)f'(1)-g(0)f'(0)-(g'(1)f(1)-g'(0)f(0))=0f'(1)-0f'(0)-g'(1)0+g'(0)0=0$

This is what we wanted yay
OH. wow. okay thank you that makes A LOT of sense. I was having an idiot moment.