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Q2). In questions from 2.1 to 2.10 below, each black (_______) is to be filled suitably.

2.6). The value of the double integral $\int^{1}_{0} \int_{0}^{\frac{1}{x}} \frac {x}{1+y^2} dxdy$ is_________.

asked in Calculus by Veteran (59.7k points) | 355 views

1 Answer

+1 vote

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.

answered by Boss (12.8k points)

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