# Integrals for exam

• Nov 20th 2006, 04:56 PM
taryn
Integrals for exam
Hey I have an exam in 2 days and I really need some help gettin these questions under control!

integral (x^2*e^-x)dx
I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2
but then I am just confused after that!

integral (2x+4)/(x^3-2x^2)dx
For this one I factorised and bottom line so I had
integral (2x+4)/(x^2)(x-2) dx

Then I sed that (2x+4)/(x^2)(x-2)= A/x+b/x+c/x-2

Then I didnt really know wat to do after that!

3. The functions S and T are defined by setting
S(x):= (integral x^2 to 0) sqrt(1+t^2)dt

I tried doin this one but when I substituted for 0 later on I got that it was undefined...

If someone could show the workings or at least the start of these problems I would be much appreciative... then I would be able to use these as examples for the other practice exams!
• Nov 20th 2006, 05:19 PM
ThePerfectHacker
Quote:

Originally Posted by taryn
Hey I have an exam in 2 days and I really need some help gettin these questions under control!

integral (x^2*e^-x)dx
I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2
but then I am just confused after that!

Integration by parts.
$\int x^2 e^{-x} dx$
Let, $u=x^2$ and $v'=e^{-x}$
Thus,
$u'=2x$ and $v=-e^{-x}$
Thus,
$uv-\int u'vdx$
Thus, (note the signs change :eek: )
$-x^2e^{-x}+2\int xe^{-x} dx$
Do the same integration by parts on this integral.
$u=x$ and $v'=e^{-x}$
Thus,
$u'=1$ and $v=-e^{-x}$
Thus, (note the parantheses :eek: )
$-x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)$
Thus,
$-x^2e^{-x}-2xe^{-x}-2e^{-x}+C$
Thus,
$-e^{-x}(x^2+2x+2)+C$
• Nov 20th 2006, 05:28 PM
taryn
Quote:

Originally Posted by ThePerfectHacker
Integration by parts.
$\int x^2 e^{-x} dx$
Let, $u=x^2$ and $v'=e^{-x}$
Thus,
$u'=2x$ and $v=-e^{-x}$
Thus,
$uv-\int u'vdx$
Thus, (note the signs change :eek: )
$-x^2e^{-x}+2\int xe^{-x} dx$
[/tex]

This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral?
$uv-\int u'vdx$

Thanks for ur help!
• Nov 20th 2006, 05:30 PM
ThePerfectHacker
Quote:

Originally Posted by taryn
This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral?
$uv-\int u'vdx$

Thanks for ur help!

I do not understand.
If you are asking whether you can do this,
$\int x^2 e^x dx=x^2\int e^x dx$
No because $x^2$ is not a constant function throughout the interval of integration.
• Nov 20th 2006, 05:41 PM
taryn
Quote:

Originally Posted by ThePerfectHacker
Integration by parts.
$\int x^2 e^{-x} dx$
Let, $u=x^2$ and $v'=e^{-x}$
Thus,
$u'=2x$ and $v=-e^{-x}$
Thus,
$uv-\int u'vdx$
Thus, (note the signs change :eek: )
$-x^2e^{-x}+2\int xe^{-x} dx$
Do the same integration by parts on this integral.
$u=x$ and $v'=e^{-x}$
Thus,
$u'=1$ and $v=-e^{-x}$
Thus, (note the parantheses :eek: )
$-x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)$
Thus,
$-x^2e^{-x}-2xe^{-x}-2e^{-x}+C$
Thus,
$-e^{-x}(x^2+2x+2)+C$

Quote:

Originally Posted by ThePerfectHacker
I do not understand.
If you are asking whether you can do this,
$\int x^2 e^x dx=x^2\int e^x dx$
No because $x^2$ is not a constant function throughout the interval of integration.

No I know thats not possible... I just mean how can u just take the uv out of the integral!
Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out?
Is there a web site or somethin that I can read to help m e understand wat u have done?
• Nov 20th 2006, 07:26 PM
ThePerfectHacker
Quote:

Originally Posted by taryn
No I know thats not possible... I just mean how can u just take the uv out of the integral!
Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out?
Is there a web site or somethin that I can read to help m e understand wat u have done?

Wikipedia explains it here.

But be warned it sometimes like to overcomplicate things (I was told I do that too).

Basically it is a special rule. You can take out that factor but then you need to change the integral. So you are not really taking the function out of the integral you are used making the integral simpler to work with.