The area (in sq. units) bounded by $y=2^x$ and $y=2x-x^2$ from $x=1$ to $x=2$ is $k{\log }_{2}e-l$, then the value of $\left|\frac{k}{l}\right|$ is equal to

Solution

Required area is
$A=\underset{1}{\overset{2}{\int }}\left[2^x-\left(2x-x^2\right)\right]dx$
$={\left[\frac{2^x}{\ln 2}-x^2+\frac{x^3}{3}\right]}_1^2$
$=\left(\frac{4}{\ln 2}-4+\frac{8}{3}\right)-\left(\frac{2}{\ln 2}-1+\frac{1}{3}\right)$
$=\frac{2}{\ln 2}-3+\frac{7}{3}$
$=\frac{2}{\ln 2}-\frac{2}{3}=2\mathrm{l}\mathrm{o}{\mathrm{g}}_{2}e-\frac{2}{3}$
$\therefore k=2 \& l=\frac{2}{3}$
$\Rightarrow \left|\frac{k}{l}\right|=3$

Hence, the correct answer is 3.