The area of the region bounded by the straight lines $x=0$ and $x=2$ and the curves $y=2^{x}$ and $\mathrm{y}=2 \mathrm{x}-\mathrm{x}^{2}$ is equal to

• A

  $\frac{2}{\log 2}-\frac{4}{3}$
• Correct answer
  B

  $\frac{3}{\log 2}-\frac{4}{3}$
• C

  $\frac{1}{\log 2}-\frac{4}{3}$
• D

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

Solution

Required area $=\int_{0}^{2}\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right) \mathrm{dx}$
$=\int_{0}^{2}\left(2^{x}-2 x+x^{2}\right) d x$ $=\left[\frac{2^{x}}{\log 2}-x^{2}+\frac{x^{3}}{3}\right]_{0}^{2}$ $=\frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}=\frac{3}{\log 2}-\frac{4}{3}$

Hence, the correct answer is option: 'B'.