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

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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- Nov 10th 2008, 06:52 AMlarsonDouble Integration --- help!
$\displaystyle \int_0^{1} \int_0^{1} \sqrt(x + y + 1)$

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. - Nov 10th 2008, 08:40 AMChris L T521

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

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

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

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

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

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

I'm sure you can take it from here. :D

--Chris - Nov 10th 2008, 08:44 AMtbgh
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.

$\displaystyle

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

$

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

$\displaystyle

\int_0^{1} \int_0^{1} (x + y + 1)^{\frac {1} {2}} dx dy

$

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.

$\displaystyle

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

$

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