Area Question 3
Question 3
- 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) $\frac{3}{2}$ (b) $\log _{e} 2+\frac{3}{2}$ (c) $\frac{1}{2}$ (2019 Main, 10 April II) (d) $\frac{3}{2}-\frac{1}{\log _{e} 2}$
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Solution:
- Given, equations of curves
$y=2^{x}$ and $y=|x+1|=\begin{array}{cc}x+1 & , x \geq-1 \ -x-1 & , x<-1\end{array}$
$\because$ The figure of above given curves is
In first quadrant, the above given curves intersect each other at $(1,2)$.
So, the required area $=\int_{0}^{1}\left((x+1)-2^{x}\right) d x$
$$ \begin{aligned} & =\frac{x^{2}}{2}+x-\frac{2^{x}}{\log {e} 2}{ }{0}^{1} \quad \because \int a^{x} d x=\frac{a^{x}}{\log _{e} a}+C \ & =\frac{1}{2}+1-\frac{2}{\log _{e} 2}+\frac{1}{\log _{e} 2} \ & =\frac{3}{2}-\frac{1}{\log _{e} 2} \end{aligned} $$
