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Math Help - Find the integral

  1. #1
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    Find the integral

    Hi there,

    \displaystyle \int (x^2-1)e^x dx

    I chose to integrate by parts. This is what I did.

    \displaystyle \int x^2 e^x dx-\int e^x dx
    1st integration \displaystyle u=x^2 ,du=2x dx ,dv=e^x dx, v=e^x
    \displaystyle (\int x^2 e^x dx = x^2 e^x -\int e^x 2xdx)-e^x +C
    2nd integration \displaystyle u=2x, du=2dx, dv=e^x dx, v= e^x
    \displaystyle \int 2x e^x dx=2xe^x -\int e^x 2dx
    My answer \displaystyle x^2 e^x +2xe^x -2e^x -e^x +C
    \displaystyle x^2 e^x +2xe^x -3e^x +C

    Is this correct?
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  2. #2
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    Quote Originally Posted by ugkwan View Post
    Hi there,

    \displaystyle \int (x^2-1)e^x dx

    I chose to integrate by parts. This is what I did.

    \displaystyle \int x^2 e^x dx-\int e^x dx
    1st integration \displaystyle u=x^2 ,du=2x dx ,dv=e^x dx, v=e^x
    \displaystyle (\int x^2 e^x dx = x^2 e^x -\int e^x 2xdx)-e^x +C
    2nd integration \displaystyle u=2x, du=2dx, dv=e^x dx, v= e^x
    \displaystyle \int 2x e^x dx=2xe^x -\int e^x 2dx
    My answer \displaystyle x^2 e^x +2xe^x -2e^x -e^x +C
    \displaystyle x^2 e^x +2xe^x -3e^x +C

    Is this correct?

    Why do you ask? Derivate and check whether you obtain the original function in the integral.

    And no, it isn't correct: you have a sign error.

    Tonio
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  3. #3
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    I see the sign error now. I needed a second opinion.
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  4. #4
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    Quote Originally Posted by ugkwan View Post
    I see the sign error now. I needed a second opinion.


    What for? To remind you that in indefinite integrals you have the tools to check whether you

    reached the correct solution or not?

    I use to say to my students: any question that can be checked in a reasonable time (the solution to a linear

    system of equations, the solution of an indefinite integral, etc.) and STILL is incorrect will be

    heavily (de)graded.

    Tonio
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