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Math Help - indefinite integral, including e

  1. #1
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    indefinite integral, including e

    <br /> <br />
\int (x^2-1)e^{x^3-3x}dx<br />
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by foodmmm View Post
    <br /> <br />
\int (x^2-1)e^{x^3-3x}dx<br />

    multiply inside the integral by 3, and divide on the outside by 3.

    \frac{1}{3} \int (3x^2 - 3) e^{x^3 -3x}

    Good luck!!
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  3. #3
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    :(

    what do i do with the e?
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  4. #4
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by foodmmm View Post
    what do i do with the e?
    Use the Chain Rule backwards:

     \int f'(x) e^{f(x)} dx = e^{f(x)}  + C

    Note that  f(x) = x^3 - 3x and  f'(x) = 3x^2 - 3 , so the antiderivative is simply
     e^{x^3 - 3x} + C

    Good luck!!
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  5. #5
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    confused

    sorry, where did the 1/3 go?
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  6. #6
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by foodmmm View Post
    sorry, where did the 1/3 go?
    ops! the 1/3 goes in the front!

    so the final answer is  \frac{1}{3} e^{x^3 - 3x}  + C
    Last edited by mr fantastic; April 14th 2009 at 06:42 AM. Reason: Removed integral sign and the dx
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  7. #7
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    lost

    so how do you get from

    <br />
\frac{1}{3} \int (3x^2 - 3) e^{x^3 -3x}<br />

    to

    <br />
\frac{1}{3} e^{x^3 - 3x}  + C<br />

    im not sure how to take those steps in between.
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  8. #8
    Senior Member apcalculus's Avatar
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    It's the Chain Rule for Integrals. If the fully derived function is inside the integral, you can take the antiderivative easily. See the the general property I listed above in the case of the exponential function with an exponent that is also a function. As long as the first derivative is inside the integral (or can be tweaked by multiplying or diving by constant as we did), then it's very simple. The antiderivative is simply the exponential function itself, plus a constant.

    Hope this helps! Good luck!!
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  9. #9
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    Quote Originally Posted by foodmmm View Post
    so how do you get from

    <br />
\frac{1}{3} \int (3x^2 - 3) e^{x^3 -3x}<br />

    to

    <br />
\frac{1}{3} e^{x^3 - 3x} + C<br />

    im not sure how to take those steps in between.
    You were given the rule in post #4.

    So either use that rule or make a substitution. The substution in this case is u = x^3 - 3x.
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