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Thread: What am I doing wrong? (indefinite integrals)

  1. #1
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    What am I doing wrong? (indefinite integrals)

    I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

    (this is a screencap of the problem, and my solution). Thanks for your help
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  2. #2
    MHF Contributor Calculus26's Avatar
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    u = x^3 +28 du = 3x^2dx

    which is where the 1/3 comes from

    integral 0f U^1/2 is 2/3U^3/2 so your coefficient is 2/9
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    Quote Originally Posted by birdz View Post
    I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

    (this is a screencap of the problem, and my solution). Thanks for your help
    \frac{1}{3} \int 3x^2 \sqrt{x^3 + 28} \, dx

    \frac{1}{3} \cdot \frac{2}{3}\left(x^3 + 28\right)^{\frac{3}{2}} + C
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    Senior Member mollymcf2009's Avatar
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    Quote Originally Posted by birdz View Post
    I've been trying to finish my hw on web assign, but can't get this problem right. I've tried reading my book, but so far as I can tell I'm doing it the way it says to. I also checked by differentiation, and I thought it was right.

    (this is a screencap of the problem, and my solution). Thanks for your help
    u = x^3 + 28

    du = 3x^2 dx

    \frac{1}{3} du = x^2 dx

    \frac{1}{3} \int \sqrt{u} du

    = \frac{1}{3} \int u^{\frac{1}{2}} du

    Then integrate:

    \frac{1}{3} \left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\right] + C

    = \frac{1}{3}\left[\frac{2}{3} u^{\frac{3}{2}}\right] + C

    Substitute back in for u, just like you did.

    You just left out the \frac{2}{3} from your integration. Try it with that, otherwise your answer is in the right form.
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    Thank you so much!
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