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Math Help - Tricky intergal... or is it?!

  1. #1
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    Question Tricky intergal... or is it?!

    I have not done calc in about 3 years and I'm currently in a calc class and I am stuck on a reveiw question:

    integrate : (x^4)e^(x^5)dx

    humm i cant seem to get it to display nice so but this is what im trying to write:

    (x^4)e^(x^5)dx

    I bet it's pretty simple, I forget alot :P Any help is greatly appriciated!
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  2. #2
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    A simple u-substitution will do.

    Let: {\color{red}u = x^5} \ \Rightarrow \ du = 5x^4 \ dx \ \Leftrightarrow \ {\color{blue}\frac{du}{5} = x^4 \ dx }

    So your integral becomes: \int {\color{blue}x^4}e^{\displaystyle {\color{red}x^5}} {\color{blue}dx} \ = \ \int e^{\displaystyle {\color{red}u}} \cdot {\color{blue}\frac{du}{5}}

    Factor out the \tfrac{1}{5} and it should be a standard integral
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  3. #3
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    Quote Originally Posted by Krooger View Post
    I have not done calc in about 3 years and I'm currently in a calc class and I am stuck on a reveiw question:

    integrate : (x^4)e^(x^5)dx

    humm i cant seem to get it to display nice so but this is what im trying to write:

    (x^4)e^(x^5)dx

    I bet it's pretty simple, I forget alot :P Any help is greatly appriciated!
    This can be solved by substitution, let u=x^5 and du = 5x^4dx or x^4dx = \frac{du}{5}

    then you can write the integral as \frac{1}{5}\int{e^udu}, integrate and substitute back in to get in terms of x again.
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  4. #4
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    Awesome thank you for the help!
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