Question

Area bounded by $y=2x-x^2$ & $(x-1{)}^2+y^2=1$ in first quadrant, is:

• A. $\frac{\pi }{2}-\frac{4}{3}$
• B. $\frac{\pi }{2}-\frac{2}{3}$
• C. $\frac{\pi }{2}+\frac{4}{3}$
• D. $\frac{\pi }{2}+\frac{2}{3}$

Answer 1

The correct option is A $\frac{\pi }{2}-\frac{4}{3}$

Area of circle is $\pi r^2$

But here only half circle is in $1^{st}$ quadrant

Then, $A^{\ast }=\frac{\pi r^2}{2}$ .......$(r=1)$

$\Rightarrow A^{\ast }=\frac{\pi }{2}......\left(1\right)$

Now area bounded by parabola $y=2x-x^2$ is

$A^{@}={\int }_0^2(2x-x^2)$

$A=\frac{4}{3}......\left(2\right)$

Now area bounded by both curves is

$A=A^{\ast }-A^{@}$

$\Rightarrow A=\frac{\pi }{2}-\frac{4}{3}$