# Use double integrals to show result of integration by parts

1. Jul 7, 2009

### theneedtoknow

1. The problem statement, all variables and given/known data

Let F(x) and G(x) be the antiderivatives of f(x) and g(x) on [a,b]. using multiple integration, show that the integral from a to b of f(x)G(x)dx = F(b)G(b)-F(a)G(a) - the integral from a to b of g(y)F(y)dy

To do so, consider the double integral of a suitable function h(x,y) over the triangle T with vertices (a,a) (b,a) (b,b)

(i'm typing up my attempt as we speak)
2. Relevant equations

3. The attempt at a solution

2. Jul 7, 2009

### theneedtoknow

I really don't have the slightest idea of how to do this...So if I consider this triangle (a, a), (b, a), (b, b), then I get a line of slope (b-a)/(b-a) = 1 joining two of the vertices. it's equation is y = x-a
So i can do a double integral with the y-bounds being y = x-a and y = a, and the x bounds a and b

(I don' tknow how to draw integrals so I will attempt to write this as coherently as possible using simple keybord commands)

The integral would be :
Integral from a to b [ Integral from y=a to y = x-a of h(x,y) dy] dx

But What can i really do with this? What would be a suitable function to replaxe h(x, y) with so i can do some actual integration, cause as of right now I have no idea how to connect that triangle, with the equation for integration by parts

3. Jul 7, 2009

### cepheid

Staff Emeritus
The equation of the line is just y = x (considering that its passes through coordinate points (a,a) and (b,b)!)

I'll try to comment further in a minute.

4. Jul 7, 2009

### theneedtoknow

ouuuu what a brainfart... thanks for pointing it out :D so the limits of integration are y=a and y=x then
argh im still trying to figure out what to do next and its frustrating haha

5. Jul 7, 2009

### cepheid

Staff Emeritus
The only thing I can think of is that you have three lines along which you can integrate and three results (the terms in the integration by parts equation) to end up with. Showing that they are related might have something to do with the relationship between the sum of the integrals along the two sides, and the integral along the hypoteneuse (I don't know, are double integrals additive)?

6. Jul 7, 2009

### Dick

Did you try the obvious sort of guess, h(x,y)=f(x)*g(y)? That's not the whole story, but if you work that one out you should be able to guess the missing ingredients of the proof.

Last edited: Jul 7, 2009
7. Jul 7, 2009

### theneedtoknow

Thank you very much for your help :)
I integrated the function you mentioned and I got something that looks very similar to what I'm trying to prove...by the presence of those functions of y on the right hand side, my guess is that the next step is to redo the integral as a type 2 instead of a type 1, and since the integral is equal no matter which way you do it, I should set the 2 results equal to each other and hopefully all the terms I need will be in there, and all the ones I don't will drop out. I'm going to try that and see how it works out :D

8. Jul 7, 2009

### Dick

You've got it. Reverse the order of integration and set the two equal. Worked for me.

9. Jul 7, 2009

### theneedtoknow

Ah, it worked out beautifully, thank you :)