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Q.
The region bounded by the curves $x=\frac{1}{2}, x=2, y=\log \,x$ and $y=2^{x}$, then the area of this region, is
Application of Integrals
Solution:
Since, the inverse of a logarithmic function is an exponential function and vice-versa and these two curves are on the opposite sides of the line $y=x$. Thus, $y=2^{x}$ and $y=\log x$ do not intersect. Their graphs are shown in figure. The shaded region in figure shows the area bounded by the given curves. Let us slice this region into vertical strips as shown in figure. For the approximating rectangle shown in figure, we have
Length $=\left(y_{1}-y_{2}\right)$, Width $=\Delta x$
Area $=\left(y_{1}-y_{2}\right) d x$
As the approximating rectangle can move horizontally
between $x=\frac{1}{2}$ and $x=2$.
$\therefore $ Required area $=\int\limits_{1 / 2}^{2}\left(y_{1}-y_{2}\right) d x=\int\limits_{1 / 2}^{2}\left(2^{x}-\log x\right) d x$
$\begin{bmatrix}\because P\left(x, y_{1}\right) \text { and } Q\left(x, y_{2}\right) \text { lie on } \\ y=2^{x} \text { and } y=\log x \text { respectively. } \\ \therefore y_{1}=2^{x} \text { and } y_{2}=\log x\end{bmatrix}$
$=\left[\frac{2^{ x }}{\log 2}- x \log x + x \right]_{1 / 2}^{2}$
$=\left\{\frac{4}{\log 2}-2 \log 2+2\right\}-\left\{\frac{\sqrt{2}}{\log 2}+\frac{1}{2} \log 2+\frac{1}{2}\right\}$
$=\frac{(4-\sqrt{2})}{\log 2}-\frac{5}{2} \log 2+\frac{3}{2} sq$ unit
