Mathematics - Area Under Curves Question with Solution | TestHub

The area of the region is given by

$A={\int }_0^2{2}^{kx}dx$

$A={\left[\frac{2^{kx}}{k\log 2}\right]}_0^2$

$A=\frac{2^{2k}-1}{k\log 2}$

Given that the area is $\frac{2}{\log 2}$:

$\frac{2^{2k}-1}{k\log 2}=\frac{2}{\log 2}$

$2^{2k}-1=2k$

Checking the given options, for $k=\frac{1}{2}$:

$2^{2(1/2)}-1=2\left(\frac{1}{2}\right)$

$2-1=1$

This satisfies the equation, hence $k=\frac{1}{2}$.

Answer: $\frac{1}{2}$