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Math Help - Double Integration --- help!

  1. #1
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    Double Integration --- help!

    \int_0^{1} \int_0^{1} \sqrt(x + y + 1)

    I'm having trouble finding the answer to this problem. The square root is supposed to be over all of the x + y + 1.

    Thanks.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by larson View Post
    \int_0^{1} \int_0^{1} \sqrt(x + y + 1)

    I'm having trouble finding the answer to this problem. The square root is supposed to be over all of the x + y + 1.

    Thanks.

    \int_0^1\int_0^1\sqrt{x+y+1}\,dy\,dx

    Note that when we integrate with respect to y, x+1 is considered a constant, let's say A.

    The integral becomes \int_0^1\int_0^1 \sqrt{y+A}\,dy\,dx

    Now to evaluate this integral, you can make a simple substitution: u=y+A\implies \,du=\,dy. The limits of integration change as well! u(0)=A and u(1)=A+1

    Thus, we now have \int_0^1\left[\int_A^{A+1}\sqrt{u}\,du\right]\,dx=\int_0^1\left[\left.\tfrac{2}{3}u^{\frac{3}{2}}\right|_A^{A+1}\r  ight]\,dx

    Now plugging back A=x+1 and simplifying, we end up with \tfrac{2}{3}\int_0^1\left[(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right]\,dx

    I'm sure you can take it from here.

    --Chris
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  3. #3
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    I'm going to assume you're integrating by x and then by y. In this case it doesn't matter if it's the opposite, but sometimes it does.

    <br />
\int_0^{1} \int_0^{1} \sqrt(x + y + 1) dx dy<br />

    The key is just remembering that the square root of a number is just an exponent of 1/2.

    <br />
\int_0^{1} \int_0^{1} (x + y + 1)^{\frac {1} {2}} dx dy<br />

    From here you're going to integrate as with any normal exponent. Add one to it and multiply the front by the inverse. Then just plug in 1 and 0 for x and subtract.

    <br />
\int_0^{1} (\frac {2} {3} * (1 + y + 1)^{\frac {3} {2}}-\frac {2} {3} * (0 + y + 1)^{\frac {3} {2}}) dy<br />

    Just repeat for the other variable and you'll be there soon enough.
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