Question

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

• A. $\left[\left(\frac{2}{\log \left(2\right)}\right)-\left(\frac{4}{3}\right)\right]$
• B. $\left[\left(\frac{3}{\log \left(2\right)}\right)-\left(\frac{4}{3}\right)\right]$
• C. $\left[\left(\frac{1}{\log \left(2\right)}\right)-\left(\frac{4}{3}\right)\right]$
• D. $\left[\left(\frac{4}{\log \left(2\right)}\right)-\left(\frac{3}{2}\right)\right]$

Answer 1

The correct option is B

$\left[\left(\frac{3}{\log \left(2\right)}\right)-\left(\frac{4}{3}\right)\right]$

Explanation for the correct option:

Given curves,

$\mathrm{y}=2^{\mathrm{x}}$ and $\mathrm{x}=0$

$\mathrm{y}=2\mathrm{x}-{\mathrm{x}}^2$ and $\mathrm{x}=2$

Required area,

$$\begin{gathered} \mathrm{A}={\int }_0^2\left(2^{\mathrm{x}}-2\mathrm{x}+{\mathrm{x}}^2\right)d\mathrm{x} \\ ={\left[\frac{2^{\mathrm{x}}}{\log \left(2\right)}-{\mathrm{x}}^2+\frac{{\mathrm{x}}^3}{3}\right]}_0^2 \\ =\frac{4}{\log \left(2\right)}-4+\frac{8}{3}-\frac{1}{\log \left(2\right)} \\ =\frac{3}{\log \left(2\right)}-\frac{4}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$

Hence, option $\left(\mathrm{B}\right)$ is the correct answer.