Question

The area (in sq. units) bounded by the curves $C_1:y=2x-x^2,x\in R$ and $C_2:y=\tan \left(\frac{\pi x}{4}\right),x\in [0,2)$ 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 $\frac{2}{3}-\frac{2\ln 2}{\pi }$


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