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Thread: definite integral

  1. #1
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    definite integral

    a little confused on this one

    "solve the definite integral"

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by JGaraffa View Post
    a little confused on this one

    "solve the definite integral"

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$
    Well, again, use substitution $\displaystyle z := x^3+5$, which gives that $\displaystyle dz=3x^2\,dx$ and hence you have

    $\displaystyle \int 2x^2e^{x^3+5}\,dx=\tfrac{2}{3}\cdot\int e^{x^3+5}\cdot 3x^2\, dx=\tfrac{2}{3}\cdot\int z\,dz=\ldots$
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  3. #3
    Junior Member piglet's Avatar
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    Quote Originally Posted by JGaraffa View Post
    a little confused on this one

    "solve the definite integral"
    $\displaystyle \int^2_0 2x^2e^{x^3+5}$
    What exactly are you confused about? Have you made an attempt at it?

    Also the integral should be $\displaystyle \int^2_0 2x^2e^{x^3+5}dx$

    I could tell you how to do it but i think i it's only fair that an attempt is made first

    Hint: use substitution
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  4. #4
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    ah perfect! i couldn't figure out how i could change that $\displaystyle 2x^2$ to a $\displaystyle 3x^2$ . thanks!
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  5. #5
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    actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$

    u= $\displaystyle x^3 + 5$
    h(u) = $\displaystyle 3x^2$

    $\displaystyle \frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

    when x = 0, u = 5
    when x = 2, u = 13

    $\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
    $\displaystyle

    \frac{2}{3} e^{13} - \frac{2}{3} e^5$

    i'm getting 294,843.32 which makes me think i'm doing something wrong.
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by JGaraffa View Post
    actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$

    u= $\displaystyle x^3 + 5$
    h(u) = $\displaystyle 3x^2$

    $\displaystyle \frac{2}{3} \int^2_0 3x^2 e^{x^3+5}$

    when x = 0, u = 5
    when x = 2, u = 13

    $\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
    $\displaystyle

    \frac{2}{3} e^{13} - \frac{2}{3} e^5$

    i'm getting 294,843.32 which makes me think i'm doing something wrong.
    No, that result is ok. Of course, $\displaystyle e^{x^3+5}$ is a fast growing function.
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  7. #7
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    Quote Originally Posted by JGaraffa View Post
    actually, i'm getting really high numbers for this which makes me think i may be doing it wrong.

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$

    u= $\displaystyle x^3 + 5$
    h(u) = $\displaystyle 3x^2$

    $\displaystyle \frac{2}{3} \int^2_0 3x^{2} e^{x^3+5}$

    when x = 0, u = 5
    when x = 2, u = 13

    $\displaystyle \frac{2}{3} \int^{13}_5 e^u du$
    $\displaystyle

    \frac{2}{3} e^{13} - \frac{2}{3} e^5$

    i'm getting 294,843.32 which makes me think i'm doing something wrong.
    Apart from [tex]3x^{2} it lookes fine to me.
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by JGaraffa View Post
    a little confused on this one

    "solve the definite integral"

    $\displaystyle \int^2_0 2x^2e^{x^3+5}$
    Again observe that $\displaystyle 3x^2$ is the derivative of $\displaystyle x^3+5$, so consider:

    $\displaystyle \frac{d}{dx}e^{x^3+5}$

    CB
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