evaluate int 0-2 [x.f(x)] where f(x)=int 2-x [e^(-t^3)]

[alex83]

evaluate: integral 0-2 [x.f(x)]

where f(x)=integral 2-x [e^(-t^3)]

i didnt get the double integrals in class yet
is there any way to solve this pls


thank u

 

[stapel]

Integrate the "inner" integral by the usual methods. Whatever result you get, plug this in for "f(x)" in the "outer" integral. Proceed as usual. :wink:

If you get stuck, please reply showing your steps and reasoning so far. Thank you!

 

[galactus]

Re: evaluate integral

Is this what you mean:

$\displaystyle \int 0-2xf(x)$

Where $\displaystyle f(x)=2-xe^{-t^{3}}$

Please don't tell me this is it:

$\displaystyle f(x)=\int_{2}^{x}e^{-t^{3}}dt$

If so, please brush up on your notation.

If the latter is it, this has no closed form and is tough to integrate.

$\displaystyle \int_{0}^{2}\int_{2}^{x}xe^{-t^{3}}dtdx$?.

May I ask where these problems are coming from?.

 

[alex83]

yes it is the second one
i know its hard and thats why im seeking help
i connot solve it as double integrals (cuz we haven't reached there yet )
i think he wants us to solve it in this form : uv- integral vdu
i dont know if im right but i tried u=x then du = dx
dv = e^(-x^3) (fondimental of calc) and cant find v

im i in the right way or not ?
thank u

 

[galactus]

Oh, he wants you to use integration by parts?.

It's still tough. It is not easily solved by parts either.

Are you sure the professor didn't give you another typo?.

You can't find v because $\displaystyle e^{-t^{3}}$ has no closed definite form.

If it is just $\displaystyle \int_{2}^{x}e^{-t^{3}}dt$, then it is probably as exercise in the second fundamental theorem of calculus.

$\displaystyle \frac{d}{dx}\int_{a}^{g(x)}f(t)=f(g(x))g'(x)$

where we would have: $\displaystyle \frac{d}{dx}\int_{2}^{x}e^{-t^{3}}dt=e^{-x^{3}}$

 

[alex83]

yes thats how it is supposed to b done but if dv= e^(-x^3) dx then how can i find v ??

 

[galactus]

As I said, you can't. It does not have a closed form. In order to find v, you must integrate dv. But $\displaystyle dv=e^{-t^{3}}$, which is not integrable.

I do not see how parts is going to help anyway. How about posting the problem exactly as it was given. Perhaps I am missing something.

 

[alex83]

As I said, you can't. It does not have a closed form. In order to find v, you must integrate dv. But $\displaystyle dv=e^{-t^{3}}$, which is not integrable.

I do not see how parts is going to help anyway. How about posting the problem exactly as it was given. Perhaps I am missing something.

\

let f(x)=integral 2-x [e^(-t^3)]

evaluate: integral 0-2 [x.f(x)]


thats the exact question

thnx

 

[galactus]

As I said on the other forum. Forget about integrating this by the means you are used to. It is not doable. Not everything is integrable. This is one of those.

I still think this is just an exercise on the 2nd FTC. Do as I and the other poster showed you. I am done with this topic.

If you don't believe me, run it through some tech and see for yourself.

$\displaystyle \int e^{-x^{3}}dx$ does not have a definite form.