IIT-JEE 1991Question 1
Sketch the curves and identify the region bounded by $x=\dfrac12$, $x=2$, $y=\ln x$ and $y=2^{x}$. Find the area of this region.
Application of Integrals
Answer$\dfrac{4-\sqrt2}{\ln 2}-\dfrac52\ln2+\dfrac32$
Solution
On $\left[\frac12,2\right]$, $2^x>\ln x$. Area $=\int_{1/2}^{2}\left(2^x-\ln x\right)dx=\frac{4-\sqrt2}{\ln2}-\left[x\ln x-x\right]_{1/2}^{2}=\frac{4-\sqrt2}{\ln2}-\frac52\ln2+\frac32$.