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Math Help - Integration by parts and substitution

  1. #1
    Senior Member polymerase's Avatar
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    Integration by parts and substitution

    How do i integrate:

    a) x^2*cos x


    b) x^2*e^-x

    If you could help me on either or both, would be great.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    How do i integrate:

    a) x^2*cos x


    b) x^2*e^-x

    If you could help me on either or both, would be great.
    In both cases use integration by parts:
    \int u dv = uv - \int v du

    For example:
    \int x^2 cos(x) dx

    Let u = x^2 \implies du = 2x dx
    Let dv = cos(x) dx \implies v = sin(x)

    So
    \int x^2 cos(x) dx = x^2 sin(x) - \int 2x sin(x) dx

    The idea is to use the "u" to reduce the power of x in front of the other function.

    Now use integration by parts on \int x sin(x) dx using the same idea.

    Problem b) is done the same way.

    -Dan
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  3. #3
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    Quote Originally Posted by polymerase View Post
    b) x^2*e^-x
     <br />
\begin{aligned}<br />
\int {x^2 e^{ - x} \,dx} &= \int {\left( {x^2 e^{ - x} + 2xe^{ - x} + 2e^{ - x} - 2xe^{ - x} - 2e^{ - x} } \right)\,dx}\\<br />
&= \int {\left[ {e^{ - x} \left( {x^2 + 2x + 2} \right) - e^{ - x} \left( {2x + 2} \right)} \right]\,dx}\\<br />
&= \int {\left[ { - e^{ - x} \left( {x^2 + 2x + 2} \right)} \right]^{\prime} \,dx} \hfill \\<br />
&= - e^{ - x} \left( {x^2 + 2x + 2} \right) +k<br />
\end{aligned}<br />
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
     <br />
\begin{aligned}<br />
\int {x^2 e^{ - x} \,dx} &= \int {\left( {x^2 e^{ - x} + 2xe^{ - x} + 2e^{ - x} - 2xe^{ - x} - 2e^{ - x} } \right)\,dx}\\<br />
&= \int {\left[ {e^{ - x} \left( {x^2 + 2x + 2} \right) - e^{ - x} \left( {2x + 2} \right)} \right]\,dx}\\<br />
&= \int {\left[ { - e^{ - x} \left( {x^2 + 2x + 2} \right)} \right]^{\prime} \,dx} \hfill \\<br />
&= - e^{ - x} \left( {x^2 + 2x + 2} \right) +k<br />
\end{aligned}<br />
    Oh that's clever! I've never seen that before.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Oh that's clever! I've never seen that before.
    Look at here

    There are another clever solutions.

    Don't be afraid about the language, 'cause that solutions don't require explanations

    Best Regards
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Krizalid View Post
    Look at here

    There are another clever solutions.

    Don't be afraid about the language, 'cause that solutions don't require explanations

    Best Regards
    Interesante. Esta bueno. Hablo un poco Espanol. Gracias!

    -Dan
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  7. #7
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    I can help ya if you wanna learn more
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  8. #8
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    I can help ya if you wanna learn more
    Krizalid I get your steps up until the 3rd. can u please explain to me how u justify from line 2 to 3. I know that 2x + 2 is the derivative of x^2+2x+2 but i just can't quite see the "proof" so to say.
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  9. #9
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    I just turned the integrand into a derivative of product functions.
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  10. #10
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    I just turned the integrand into a derivative of product functions.
    Sorry to bother u again but...I still dont understand what u mean
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  11. #11
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    As you can see, in the first line I added and subtracted some terms properly.

    In the second one I factorised properly. In the third one I used [f(x)\cdot g(x)]'=f'(x)\cdot g(x)+f(x)\cdot g'(x), finally I just integrated.
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  12. #12
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Krizalid View Post
    As you can see, in the first line I added and subtracted some terms properly.

    In the second one I factorised properly. In the third one I used [f(x)\cdot g(x)]'=f'(x)\cdot g(x)+f(x)\cdot g'(x), finally I just integrated.
    Thank you i see what u did now

    PS are u doing a math major?
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  13. #13
    Math Engineering Student
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    Quote Originally Posted by polymerase View Post
    PS are u doing a math major?
    What's a math major?

    Sorry, it's just I'm chilean
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  14. #14
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    Quote Originally Posted by Krizalid View Post
    What's a math major?
    When you go to University you select an area what you want to study. Like physics, chemistry, .... When someone says "what is your major" he means what area you selected.


    By the way, you site has the coolest graphics, I would join but I do not speak Spanish.
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  15. #15
    Math Engineering Student
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    Ahh ok, now I get it.

    Hey Hacker, you'd really be a great value if you register there, 'cause you're a really good mathematician.

    Actually you can use Google Language Tools to translate.

    Anyway if you desire, I can help you with spanish so you can post there

    P.S.: I'm not in the college yet, so I didn't select a math major
    Last edited by Krizalid; October 14th 2010 at 06:30 PM.
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