• question_answer Using integration, find the area of the following region. $\{(x,\,\,y):|x-1|\,\,\le \,y\le \sqrt{5-{{x}^{2}}}\}$

Given region is $\{(x,\,\,y):|x-1|\le y\le \sqrt{5-{{x}^{2}}}\}$ Above region has two equations $y=\,|x-1|$and $y=\sqrt{5-{{x}^{2}}}$ Now, we know that, |x|\,\,=\left\{ \begin{align} & \,\,\,\,x,\,\,\,if\,\,x\ge 0 \\ & -\,x,\,\,\,if\,\,x<0 \\ \end{align} \right. Using above definition of modulus functions, we write y=\,\,|x-1|\,\,=\left\{ \begin{align} & \,\,\,\,\,\,x-1,\,\,\,if\,\,x-1\ge 0 \\ & -\,(x-1),\,\,\,if\,\,x-1<0 \\ \end{align} \right. y=\left\{ \begin{align} & x-1,\,\,\,if\,\,x\ge 1 \\ & 1-x,\,\,\,if\,\,x<1 \\ \end{align} \right. Also, other equation is$y=\sqrt{5-{{x}^{2}}}$ On squaring both sides, we get ${{y}^{2}}=5-{{x}^{2}}\Rightarrow {{x}^{2}}+{{y}^{2}}=5$ It is a circle with centre (0, 0) and radius, $r=\sqrt{5}$. On drawing the rough sketch, we get the following graph For finding the points of intersection of the curves, we have $y=1-x$    ?(i) $y=x-1$             ?(ii) and ${{x}^{2}}+{{y}^{2}}=5$ On putting $y=1-x$from Eq. (i) in Eq. (iii), we get ${{x}^{2}}+{{(1-x)}^{2}}=5$ $\Rightarrow$   ${{x}^{2}}+1+{{x}^{2}}-2x=5$ $\Rightarrow$   $2{{x}^{2}}-2x-4=0$ $\Rightarrow$   ${{x}^{2}}-x-2=0$ $\Rightarrow$   ${{x}^{2}}-2x+x-2=0$ $\Rightarrow$$x\,(x-2)+1\,(x-2)=0$ $\Rightarrow$   $(x+1)\,(x-2)=0$ $\therefore$                  $x=-\,1$or 2 Now, when$x=-\,1$, then ${{y}^{2}}=5-{{x}^{2}}=5-1=4$ $\Rightarrow$               ${{y}^{2}}=4$          $\Rightarrow$$y=\pm \,\,2$ and when $x=2$, then ${{y}^{2}}=5-{{x}^{2}}=5-4=1$ $\Rightarrow$               ${{y}^{2}}=1$ $\Rightarrow$               $y=\pm \,\,1$ So, points of intersection of Eqs. (i) and (ii) are $(-\,1,\,\,\pm \,\,2)$and $(2,\,\,\pm \,\,1)$. Now, putting $y=x-1$from Eq. (ii) in Eq. (iii), we get ${{x}^{2}}+{{(x-1)}^{2}}=5$$\Rightarrow$${{x}^{2}}+{{x}^{2}}+1-2x=5$ $\Rightarrow$$2{{x}^{2}}-2x-4=0$$\Rightarrow$${{x}^{2}}-x-2=0$ $\Rightarrow$$(x-2)\,(x+1)=0$$\Rightarrow$$x=-\,1$or 2 From Eq. (iii), at$x=-\,1$, $y=\pm 2$and $x=2,$$y=\pm 1$Hence, the two curves intersect at$(-\,1,\,\,\pm \,\,2)$and $(2,\,\,\pm \,\,1)$. Now, required area $\int_{-\,1}^{2}{{{y}_{(\,\,circle\,\,)}}dx-\int_{-\,1}^{1}{(y=1-x)\,dx-\int_{1}^{2}{(y=x-1)\,dx}}}$ $=\int_{-\,1}^{2}{\sqrt{5-{{x}^{2}}}dx-}\int_{-\,1}^{1}{(1-x)\,dx-\int_{1}^{2}{(x-1)\,dx}}$ $\left[ \because \int{\sqrt{{{a}^{2}}-{{x}^{2}}}dx}=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-\,1}}\frac{x}{a} \right]$ $=\left[ \frac{x}{2}\sqrt{5-{{x}^{2}}}+\frac{5}{2}{{\sin }^{-\,1}}\frac{x}{\sqrt{5}} \right]_{-\,1}^{2}-\left[ x-\frac{{{x}^{2}}}{2} \right]_{-\,1}^{1}-\left[ \frac{{{x}^{2}}}{2}-x \right]_{1}^{2}$$=\left[ \left( \frac{2}{5}\sqrt{5-4}+\frac{5}{2}{{\sin }^{-\,1}}\frac{2}{\sqrt{5}} \right) \right]$ $\left. -\left\{ -\frac{1}{2}\sqrt{4}+\frac{5}{2}{{\sin }^{-\,1}}\left( -\frac{1}{\sqrt{5}} \right) \right\} \right]$ $-\,\left[ \left( 1-\frac{1}{2} \right)-\left( -1-\frac{1}{2} \right) \right]-\left[ \left( \frac{4}{2}-2 \right)-\left( \frac{1}{2}-1 \right) \right]$ $=1+\frac{5}{2}{{\sin }^{-1}}\frac{2}{\sqrt{5}}+\frac{1}{2}\times 2-\frac{5}{2}{{\sin }^{-\,1}}\left( -\frac{1}{\sqrt{5}} \right)$ $-\left( \frac{1}{2}+\frac{3}{2} \right)-\left( 0+\frac{1}{2} \right)$ $=1+\frac{5}{2}{{\sin }^{-1}}\frac{2}{\sqrt{5}}+1+\frac{5}{2}{{\sin }^{-\,1}}\frac{1}{\sqrt{5}}-2-\frac{1}{2}$ $[\because {{\sin }^{-\,1}}(-\,\theta )=-{{\sin }^{-\,1}}\theta ]$ $=-\frac{1}{2}+\frac{5}{2}\left( {{\sin }^{-\,1}}\frac{2}{\sqrt{5}}+{{\sin }^{-\,1}}\frac{1}{\sqrt{5}} \right)$ $=\left[ \frac{5}{2}\left( {{\sin }^{-1}}\frac{2}{\sqrt{5}}+{{\sin }^{-\,1}}\frac{1}{\sqrt{5}} \right)-\frac{1}{2} \right]$ $=\frac{5}{2}\left[ {{\cos }^{-\,1}}\frac{1}{\sqrt{5}}+{{\sin }^{-\,1}}\frac{1}{\sqrt{5}} \right]-\frac{1}{2}$ $\left[ \because {{\sin }^{-\,1}}\frac{2}{\sqrt{5}}={{\cos }^{-\,1}}\frac{1}{\sqrt{5}} \right]$ $\frac{5}{2}\times \frac{\pi }{2}-\frac{1}{2}=\frac{1}{4}(5\pi -2)$ sq units