Question

Area of the region bounded by the curves $y=2^x,y=2x-x^2,x=0$ and $x=2$ is given by

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

Answer 1

Answer: (A)

$\frac{3}{\log 2}-\frac{4}{3}$

Let the required area be $A$ sq. units. Then,
$A={\int }_0^2(y_2-y_1)dx$
$={\int }_0^2(2^x-(2x-x^2\left)\right)dx$
$={[\frac{2^x}{\log 2}-x^2+\frac{x^3}{3}]}_0^2$
$=\frac{4}{\log 2}-4+\frac{8}{3}\frac{1}{\log 2}$
$=\frac{3}{\log 2}-\frac{4}{3}sq.units$