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Math Help - Integral involving e, exponents and fractions

  1. #1
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    Integral involving e, exponents and fractions

    Is someone able to show me how I'd calculate this integral (showing working steps)? Any help would be greatly appreciate.

    \int{{-5x^3 e^{-5x}}\over{3}}  - {5x^2 e^{-5x} dx}
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by drew.walker View Post
    Is someone able to show me how I'd calculate this integral (showing working steps)? Any help would be greatly appreciate.

    \int{{-5x^3 e^{-5x}}\over{3}} - {5x^2 e^{-5x} dx}
    This is the same as

    \tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx=\tfrac{1}{3}\int\frac{\,d}{\,dx}\left[x^3e^{-5x}\right]\,dx-6\int x^2e^{-5x}\,dx =\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx.

    Now apply integration by parts twice to \int x^2e^{-5x}\,dx and then you're done!

    Can you take it from here?
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  3. #3
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    Yeah, I think I'm okay from there. I knew there was a way to simplify the integral, but I just couldn't see it. Been doing this for too long tonight! Thanks.
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  4. #4
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    Actually, something I'm confused about. Why did you bother making the equation:

    <br />
\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx<br />

    rather than just:
    <br />
\tfrac{1}{3}\int-5x^3e^{-5x}\,dx-\int5x^2e^{-5x}\,dx<br />

    Wouldn't the second option be equal, but easier to solve?
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by drew.walker View Post
    Actually, something I'm confused about. Why did you bother making the equation:

    <br />
\tfrac{1}{3}\int\left[-5x^3e^{-5x}+3x^2e^{-5x}\right]\,dx-\int6x^2e^{-5x}\,dx<br />

    rather than just:
    <br />
\tfrac{1}{3}\int-5x^3e^{-5x}\,dx-\int5x^2e^{-5x}\,dx<br />

    Wouldn't the second option be equal, but easier to solve?
    Let's see!

    Let's focus on \int -5x^3e^{-5x}\,dx.

    Let u=-5x^3 and \,dv=e^{-5x}\,dx

    Then \,du=-15x^2\,dx and v=-\tfrac{1}{5}e^{-5x}

    Thus, \int -5x^3e^{-5x}\,dx=x^3e^{-5x}-3\int x^2e^{-5x}\,dx.

    Now, note that \tfrac{1}{3}\int -5x^3e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}\left[x^3e^{-5x}-3\int x^2e^{-5x}\,dx\right]-5\int x^2e^{-5x}\,dx =\tfrac{1}{3}x^3e^{-5x}-\int x^2e^{-5x}\,dx-5\int x^2e^{-5x}\,dx=\tfrac{1}{3}x^3e^{-5x}-6\int x^2e^{-5x}\,dx

    So you'll end up with the same thing!

    I just thought forcing product rule in the integrand would save time in integrating ==> you only need to apply integration by parts twice.
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