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Math Help - definite integral

  1. #1
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    definite integral

    a little confused on this one

    "solve the definite integral"

    \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"

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

    \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"
    \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 \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 2x^2 to a 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.

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

    u= x^3 + 5
    h(u) = 3x^2

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

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

    \frac{2}{3} \int^{13}_5 e^u du
    <br /> <br />
\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.

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

    u= x^3 + 5
    h(u) = 3x^2

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

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

    \frac{2}{3} \int^{13}_5 e^u du
    <br /> <br />
\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, 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.

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

    u= x^3 + 5
    h(u) = 3x^2

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

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

    \frac{2}{3} \int^{13}_5 e^u du
    <br /> <br />
\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"

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

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

    CB
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