Question

The area (in sq. units) of the region bounded by the curves $y=2^x$ and $y=|x+1|,$ in the first quadrant is :

• A. ${\log }_{e}2+\frac{3}{2}$
• B. $\frac{3}{2}-\frac{1}{{\log }_{e}2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{3}{2}-\frac{1}{{\log }_{e}2}$

Given ,

$y=2^x$ and $y=|x+1|=\{\begin{array}{cc}x+1,& x\ge -1\\ -x-1,& x<-1\end{array}$

$(1,2)$ is the point of intersection

Area enclosed $=\underset{0}{\overset{1}{\int }}((x+1)-2^x)dx$
$={[\frac{x^2}{2}+x-\frac{2^x}{{\log }_{e}2}]}_0^1$
$=(\frac{1}{2}+1-\frac{2}{{\log }_{e}2}+\frac{1}{{\log }_{e}2})$
$=\frac{3}{2}-\frac{1}{{\log }_{e}2}\text{sq. units}$