Question

If the area bounded by the curve $y=2^{kx},k>0$ and $x=0$, $x=2$ and $x-$ axis is $\frac{3}{\ln 2},$ then the value of $k$ is

Answer 1

The given curve is $y=2^{kx},k>0$
Clearly, $y>0$ for all values of $x$ and $k$
$\therefore$ the curve $y=2^{2k}$ lies above $x-$axis for $0\le x\le 2.$
Area bounded by the curve $y=2^{kx},x=0,x=2$ is given by:
$A=\underset{0}{\overset{2}{\int }}ydx=\underset{0}{\overset{2}{\int }}{2}^{kx}dx$
$=\frac{1}{k}\underset{0}{\overset{2k}{\int }}{2}^{t}dt,$ where $t=kx$
$=\frac{1}{k}{\left[\frac{2^t}{\ln 2}\right]}_0^{2k}=\frac{1}{k\ln 2}(2^{2k}-1)$
Now, $A=\frac{3}{\ln 2}$(given)
$\therefore \frac{1}{k\ln 2}(2^{2k}-1)=\frac{3}{\ln 2}\Rightarrow 2^{2k}-1=3k$
$\Rightarrow k=1$