Question

The area x-axis, bounded by the curve y = 2^kx and x = 0 and x = 2 is 3 log₂e, then 2^2k - 3k = ______________.

Answer 1

Given: area of the region bounded by the curve y = 2^kx and x = 0 and x = 2 is 3 log₂e

We know,
the required area of the region = ${\int }_0^{2}ydx$
Thus,
$$\begin{gathered} 3{\log }_{2}e=\left|{\int }_0^2\left(2^{kx}\right)dx\right| \\ \Rightarrow \pm 3{\log }_{2}e={\left(\frac{2^{kx}}{k\log 2}\right)}_0^2 \\ \Rightarrow \pm \frac{3}{\log 2}=\frac{1}{k\log 2}{\left(2^{kx}\right)}_0^2 \\ \Rightarrow \pm 3k={\left(2^{kx}\right)}_0^2 \\ \Rightarrow \pm 3k=\left(2^{2k}-2^0\right) \\ \Rightarrow 3k=\left(2^{2k}-1\right)\mathrm{or}-3k=\left(2^{2k}-1\right) \\ \Rightarrow 1=2^{2k}-3k\mathrm{or}1=2^{2k}+3k \\ \Rightarrow 2^{2k}-3k=1 \\ \mathrm{Thus},2^{2k}-3k=1 \end{gathered}$$


Hence, 2^2k − 3k = 1.