# Math Help - Integration by parts

1. ## Integration by parts

Hello. I've been using all the time primitivization by parts in the following way:

P (f g) = (P f)g - P ((P g) g')

Which I thought that would correctly translate to integration by parts like the following:

It seems that in the end of the expression, instead of integration f(x) from a to b, I should just let it be f(x), as it is already inside an integral. But I don't get why. Could anyone help? Thanks

2. Originally Posted by devouredelysium
Hello. I've been using all the time primitivization by parts in the following way:

P (f g) = (P f)g - P ((P g) g')

Which I thought that would correctly translate to integration by parts like the following:

It seems that in the end of the expression, instead of integration f(x) from a to b, I should just let it be f(x), as it is already inside an integral. But I don't get why. Could anyone help? Thanks

It perhaps will help if you write down the rule for integration by parts in the following way:

$\int\limits_a^b\! f(x)g(x)\,dx$ $=g(x)F(x) \mid_a^b -\int\limits_a^b\! g'(x)F(x)\,dx$ , where $F'(x)=f(x)$

Writing F for the primitive of f makes things clearer imo, and now it's a standard exercise to show the derivative of the RH equals the function under the integral sign in the LH.

Tonio

3. Originally Posted by devouredelysium
Hello. I've been using all the time primitivization by parts in the following way:

P (f g) = (P f)g - P ((P g) g')

Which I thought that would correctly translate to integration by parts like the following:

It seems that in the end of the expression, instead of integration f(x) from a to b, I should just let it be f(x), as it is already inside an integral. But I don't get why. Could anyone help? Thanks
I had a little trouble with the word "primitivization" but then recognized it as "find the primitive", i.e. the "anti-derivative".

Yes, that just expresses "integration by parts", although you have written it incorrectly: it should be P(fg)= (Pf)g- P((Pf)g').

The formula for integration by parts that I learned is $\int u dv= uv- \int v du$

Here, take u= g(x), dv= f(x)dx. Then du= g' and [/tex]v= \int f(x)dx[/tex] which, in your notation, would be "Pf".

Now, uv= g(Pf) and vdu= (Pf)g' so $uv- \int v du$ is g(Pf)- P(Pf)g' which, in "normal notation" would be
$\left(\int f(x)dx\right)g(x)- \int\left(\int f(x)dx\right) g'(x) dx$

Yes, of course that "integral" inside the outer integral should be an antiderivative or "indefinite integral". It might be best to write this
$\int_a^b f(x)g(x)dx= \left(\int_a^b f(x)dx\right)(g(b)- g(a))- \int_a^b \left(\int_0^x f(t)dt\right)g'(x)dx$

4. Yes, what I meant was P(fg)= (Pf)g- P((Pf)g'), not what I wrote. It was a typo. But converting this directly to integral form would yield what I put up there. I can see that tonio's formula is pretty similar to mine, being that the only difference I can spot is that he has in the end the anti-derivative of f(x)(which he calls F(x)) instead of an integral of f(x). And that is my question, why is it?

5. Originally Posted by devouredelysium
Yes, what I meant was P(fg)= (Pf)g- P((Pf)g'), not what I wrote. It was a typo. But converting this directly to integral form would yield what I put up there. I can see that tonio's formula is pretty similar to mine, being that the only difference I can spot is that he has in the end the anti-derivative of f(x)(which he calls F(x)) instead of an integral of f(x). And that is my question, why is it?
He just wanted to avoid the question of limits of integration!

If F(x) is an anti-derivative of f(x) then \int_a^b f(x)dx= f(b)- f(a), not what is wanted here. Of course, any anti-derivative can be written as F(x)+ C for some constant C. Notice that [tex]\int_a^x f(t)dt= F(x)- F(a) which is just F(x)+ C with C chosen to be -F(a).

Since, here, we just need an anti-derivative, we can write it as $F(x)= \int_0^x f(t)dt$ though it really doesn't matter what lower limit we use.