# Sketch the region and find the area bounded between $y=\sqrt {5-x^2}$ and $y=|x-1|$ using integration.

$\begin{array}{1 1} \large\frac{3\pi+2}{4} \\\large\frac{3\pi-2}{4} \\ \large\frac{5\pi-2}{4} \\ \large\frac{5\pi+2}{4} \end{array}$

• The area enclosed by a curve $y=f(x)$,the $x$-axis and the ordinate $x=a$ and $y=b$ is given by $\int_a^b ydx.$
• $\int\sqrt{a^2-x^2}dx=\large\frac{x}{2}$$\sqrt{a^2-x^2}+\large\frac{a^2}{2}$$\sin^{-1}\big(\large\frac{x}{a}\big)$$+c. • sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2} Step 1: Given curves are y=\sqrt {5-x^2}...........(i) and y=|x-1|.....(ii) Clearly the equation y=\sqrt {5-x^2} represents a semi circle of radius \sqrt 5 and having centre (0,0) which is above x\:axis. Step 2: y=|x-1| can be written as y=x-1........(iii) when x>1 and y=1-x...........(iv) when x<1 Clearly this represents a broken line broken at (1,0) Step 3 Solving the equations (i) and (iii) we get the point of intersection B Solving y^2=5-x^2 and y=x-1 we get \Rightarrow\:x^2+(x-1)^2=5 \Rightarrow\:x=2 and y=1 B(2,1) Similarly by solving the equations (i) and (iv) we get the point A \Rightarrow\:x^2+(1-x)^2=5\:\:\Rightarrow\:x=-1,\:\:y=2 A(-1,2) Step 4 From the figure the area of the shaded region is given by A=\int _{-1} ^{2}\sqrt {5-x^2}\:dx-\int _{-1} ^{1}(1-x)\:dx-\int _{1}^{2}(x-1)\:dx On integrating A=\begin{bmatrix}\large\frac{x}{2}\normalsize\sqrt{5-x^2}+\large\frac{5}{2}\normalsize\sin^{-1}\big(\large\frac{x}{\sqrt 5}\big)\end{bmatrix}_{-1}^{2}-\begin{bmatrix}x-\large\frac{x^2}{2}\end{bmatrix}_{-1}^1-\begin{bmatrix}\large\frac{x^2}{2}-x\end{bmatrix}_{1}^2 On applying the limits 1+\large\frac{5}{2}$$sin^{-1}(\large\frac{2}{\sqrt 5}$$)+1-\large\frac{5}{2}$$sin^{-1}(\large\frac{-1}{\sqrt 5}$$)-1+\large\frac{1}{2}$$-1-\large\frac{1}{2}$$-2+2+\large\frac{1}{2}$$-1$
$\Rightarrow\:A=\large\frac{5}{2}$$(sin^{-1}\large\frac{2}{\sqrt 5}$$+sin^{-1}\large\frac{1}{\sqrt 5})-\large\frac{1}{2}$
But $sin^{-1}\large\frac{1}{\sqrt 5}$$=cos^{-1}\large\frac{2}{\sqrt 5} \Rightarrow\:A=\large\frac{5}{2}$$(sin^{-1}\large\frac{2}{\sqrt 5}$$+cos^{-1}\large\frac{2}{\sqrt 5})-\large\frac{1}{2} Also we know that sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2} \Rightarrow\:\large\frac{5}{2}.\large\frac{\pi}{2}-\frac{1}{2}=\large\frac{5\pi-2}{4}$$sq.units.$