Question

The area (in sq. units) bounded by the curves $C_1:y=2x-x^2,x\in R$ and $C_2:y=\tan \left(\frac{\pi }{4}x\right),x\in [0,2)$ is equal to

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

Answer 1

The correct option is C $\frac{2}{3}-\frac{2\ln 2}{\pi }$

Given that,
$C_1:y=2x-x^2$ and
$C_2:y=\tan \left(\frac{\pi }{4}x\right)$
Points of intersection can be obtained by equating $C_1$ and $C_2.$
$2x-x^2=\tan \left(\frac{\pi }{4}x\right)\cdots \left(1\right)$
L.H.S. is an algebraic function and R.H.S. is a trigonometric function. So, it can not be solved by proper method. Hence we are going to use hit and trial method.
Putting $x=0$ in equation $\left(1\right)$
L.H.S. $=$ R.H.S. $=0$
Therefore, $x=0$ is a solution.
Similarly, putting $x=1$ in equation $\left(1\right)$
L.H.S. $=$ R.H.S. $=1$
So, $x=1$ is also the solution.
$\therefore (0,0)$ and $(1,1)$ are the intersection points.


Area $=\underset{0}{\overset{1}{\int }}(2x-x^2-\left(\tan \frac{\pi }{4}x\right))dx$
$={[x^2-\frac{x^3}{3}-\frac{\ln \left(\sec \frac{\pi }{4}x\right)}{\frac{\pi }{4}}]}_0^1$
$=1-\frac{1}{3}-\frac{\ln \sqrt{2}}{\frac{\pi }{4}}$
$=\frac{2}{3}-\frac{2\ln 2}{\pi }$