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Math Help - Check my work so far with this double integral problem, please?

  1. #1
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    Check my work so far with this double integral problem, please?

    The problem is to compute the double integral sin(y^2)dydx from 3x<=y<=12, 0<=x<=4.

    I wanted to check to see if I was on the right track in solving this. I figured I should switch the order of integration, so that it becomes sin(y^2)dxdy from 0<=y<=12, 0<=x<=y/3. And since sin(y^2) is constant with respect to x, then the double integral would be the same as a single integral of sin(y^2)*(y/3) dy from 0<=y<=12.

    Is this correct?
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  2. #2
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    Yes, and now you can evaluate that single integral using a u substitution.
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  3. #3
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    Got it! Thanks.
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