Question

Q. 7${\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}}\frac{x}{y+1}dxdy$

Short answer

${\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}}\frac{x}{y+1}dxdy=0$

Step-by-step solution

Draw the region

The region determined by the limits of the given integral is shown below,

Evaluate integral

Here a concept is used,

${\int }_{-a}^{a}f\left(x\right)dx=0,\mathrm{If}f\left(x\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}\mathrm{i}.\mathrm{e}.f(-x)=-f\left(x\right)$

$$\begin{gathered} {\int }_0^1{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}}\frac{x}{y+1}dxdy={\int }_0^1\frac{1}{y+1}\left[{\int }_{-\sqrt{1-y^2}}^{\sqrt{1+y^2}}xdx\right]dy \\ ={\int }_0^1\frac{1}{y+1}\left[0\right]dy\left[f\left(x\right)=x,\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}\right] \\ =0 \end{gathered}$$