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Math Help - double integral

  1. #1
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    double integral

    Hi, can someone please help me.. coz what im doing looks wrong..

     \int_0^9 \int_0^{\sqrt{x}} y cos x^2 dy dx
     cos x^2 \int_0^9 [\frac{y^2}{2}]_0^{\sqrt{x}} dx
     \int_0^9 \frac{x}{2} cos x^2 dx
     \frac{1}{2} \int_0^9 x cos x^2 dx

    let u =  x^2
    du = 2x dx
     \frac{1}{2} du = x dx

    =  \frac{1}{4} \int_0^9 cos u du
    =  \frac{1}{4} - sin u ]_0^9
    = - \frac{1}{4} sin 9

    sorry im so dumb please help me
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Adre5 View Post
    Hi, can someone please help me.. coz what im doing looks wrong..

     \int_0^9 \int_0^{\sqrt{x}} y cos x^2 dy dx
     cos x^2 \int_0^9 [\frac{y^2}{2}]_0^{\sqrt{x}} dx
     \int_0^9 \frac{x}{2} cos x^2 dx
     \frac{1}{2} \int_0^9 x cos x^2 dx

    let u =  x^2
    du = 2x dx
     \frac{1}{2} du = x dx

    =  \frac{1}{4} \int_0^9 cos u du
    =  \frac{1}{4} - sin u ]_0^9
    = - \frac{1}{4} sin 9

    sorry im so dumb please help me
     \int_0^9 \int_0^{\sqrt{x}} y \times cos x^2 dy dx

    =  \int_0^9 \int_0^{\sqrt{x}} \frac{y^2}{2} \times cos x^2 dx

    =  \int_0^9 cos x^2 \times \frac{x}{2} dx

    let u = x^2 \rightarrow du = 2x dx \rightarrow dx = \frac{du}{2x}

    the integral then becomes:

    =  \int_0^9 cos u \times \frac{x}{2} \frac{du}{2x}

    =  \frac{1}{4}\int_0^9 cos u du

    = \frac{1}{4} [sin(u)]_0^9

    = \frac{1}{4} [sin(x^2)]_0^9

    = \frac {sin(81)}{4}

    You did not substitute u = x^2 in the penultimate step of your solution!
    Last edited by harish21; April 5th 2010 at 12:45 AM.
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