Question

The area of the region bounded by the curves $y=2^x,y=2x-x^2$ and $x=2$ is

• A. $\frac{3}{\log 2}-\frac{4}{3}$
• B. $\frac{3}{\log 2}-\frac{4}{9}$
• C. $\frac{3}{2}-\frac{\log 2}{9}$
• D. None of these

Answer 1

Answer: (A)

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

Given curves are $y=2^x,y=2x-x^2$ and $x=2$
$\therefore$ Required area $={\int }_0^2[2^x-(2x-x^2\left)\right]dx$
$={\int }_0^2(2^x-2x+x^2)dx$
$={[\frac{2^x}{\log 2}-x^2+\frac{x^3}{3}]}_0^2$
$=\frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}$
$=\frac{3}{\log 2}-\frac{4}{3}$.