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The value of the double integral $\int^{1}_{0} \int_{0}^{\frac{1}{x}} \frac {x}{1+y^2} dxdy$ is_________.
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Limit of inner integral are from $0$ to $1/x$. Clearly these are the limits of $y$. So first we integrate with respect to $y$ taking $x$ as constant. After that we will integrate w.r.t $x$.

$\displaystyle\int_{0}^1\int_0^{1/x}\frac{x}{1+y^2}dx dy$

$=\displaystyle\int_{0}^1x\left(\int_0^{1/x}\frac{1}{1+y^2}dy\right) dx$

$=\displaystyle\int_{0}^1x\left(\left[tan^{-1}y\right]_0^{1/x}\right) dx$

$=\displaystyle\int_{0}^{1}x\tan^{-1}\left(\frac{1}{x}\right)dx$

$=\displaystyle\left[\tan^{-1}\left(\frac{1}{x}\right)\cdot\frac{x^{2}}{2}-\frac{1}{2} \int \frac{x^{4}}{x^{2}+1}\left(\frac{-1}{x^2}\right)dx \right]_{0}^{1}$

$=\displaystyle\left[\tan^{-1}\left(\frac{1}{x}\right)\cdot\frac{x^{2}}{2}\right]_0^1+\frac{1}{2}\left[ \int \frac{x^2}{x^{2}+1}dx \right]_{0}^{1}$

$\displaystyle=\frac{\Pi }{8}+\frac{1}{2}\left[ \int \frac{(x^2+1)-1}{x^{2}+1}dx \right]_{0}^{1}$

$\displaystyle=\frac{\Pi}{8}+ \frac{1}{2}\left[x-tan^{-1}x \right]_{0}^{1}$

$=\frac{\Pi}{8}+\frac{1}{2}-\frac{\Pi}{8}$

$=1/2$

Hence, answer should be $1/2$.
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In a double integral, the outer limits must be constant, but the inner limits can depend on the outer variable. This means, we must put y as the inner integration variables.

$\int_{0}^{1}\int_{0}^{1/x}x/1+y^{2}dxdy$

$=\int_{0}^{1}x\tan^{-1}(y)]_{0}^{1/x}dx$

$=\int_{0}^{1}x(\tan^{-1}(1/x)-\tan^{-1}(0))dx$

$=\int_{0}^{1}x\tan^{-1}(1/x)dx$

$=[\tan^{-1}(1/x)\frac{x^{2}}{2}-1/2[ \int \frac{x^{2}}{x^{2}+1} x^{2} dx ]]_{0}^{1}$

$=[\tan^{-1}(1/x)\frac{x^{2}}{2}-1/2[\frac{x^{3}}{3}-x+\tan^{-1}(x)]]_{0}^{1}$

$=(\frac{1}{2}*\frac{\Pi }{4}-1/2[1/3-1+\frac{\Pi }{4}])$

$=\frac{\Pi }{8}-1/2[-2/3+\frac{\Pi }{4}$

$=\frac{\Pi }{8}+\frac{1}{3}-\frac{\Pi }{8}$

$=1/3$

Hence, answer should be $1/3.$

4 Comments

No, firstly, we have to derivate w.r.t. x, not y.

I think this ans is not correct.
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Is multiple integrals in current syllabus?
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yes it is in syllabus.........
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