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I don't know how to text in the limits but the interior areand
, and the exterior are 0 and 2.
What do I do with this please?
This is completely correct. However, what you wrote before:Originally Posted by chisigma
Yes... without any hassle!... You have to consider that the integral is...
(1)
... where...
(2)
which would be likely to be interpreted as the product of two integrals:
which is equal to
A better notation would be
 xdx)
While what ChiSigma said is correct, it's only valid to separate the variables if the variables can be separated, as they can be in this case. A better way in my opinion, is to perform the innermost integration of the ENTIRE function, then the outer integration of the entire function. This will work in ALL cases.
I don't know how to text in the limits but the interior are
What do I do with this please?
So in your case...
![\displaystyle \begin{align*} \int_0^2{\int_{\sqrt{x}}^x{x\,y\,dy\,dx}} &= \int_0^2{\left(\int_{\sqrt{x}}^x{x\,y\,dy}\right) \,dx} \\ &= \int_0^2{\left[\frac{x\,y^2}{2}\right]_{\sqrt{x}}^x\,dx} \\ &= \int_0^2{\left[\frac{x(x)^2}{2} - \frac{x(\sqrt{x})^2}{2}\right]\,dx} \\ &= \int_0^2{\frac{x^3}{2} - \frac{x^2}{2}\,dx} \\ &= \left[\frac{x^4}{8}- \frac{x^3}{6}\right]_0^2 \\ &= \left(\frac{2^4}{8} - \frac{2^3}{6}\right) - \left(\frac{0^4}{8} - \frac{0^3}{6}\right) \\ &= \frac{2}{3}\end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \int_0^2{\int_{\sqrt{x}}^x{x\,y\,dy\,dx}} &= \int_0^2{\left(\int_{\sqrt{x}}^x{x\,y\,dy}\right) \,dx} \\ &= \int_0^2{\left[\frac{x\,y^2}{2}\right]_{\sqrt{x}}^x\,dx} \\ &= \int_0^2{\left[\frac{x(x)^2}{2} - \frac{x(\sqrt{x})^2}{2}\right]\,dx} \\ &= \int_0^2{\frac{x^3}{2} - \frac{x^2}{2}\,dx} \\ &= \left[\frac{x^4}{8}- \frac{x^3}{6}\right]_0^2 \\ &= \left(\frac{2^4}{8} - \frac{2^3}{6}\right) - \left(\frac{0^4}{8} - \frac{0^3}{6}\right) \\ &= \frac{2}{3}\end{align*} )
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