Mathematics - Area Under Curves Question with Solution | TestHub

We know that $\left|x\right|=\left\{\begin{array}{ll} x,& x\ge 0\\ -x,& x<0\end{array}\right.$

Hence, $\left|x+1\right|=\left\{\begin{array}{ll} x+1,& x\ge -1\\ -\left(x+1\right),& x<-1\end{array}\right.$
The graph of the given functions is

The shaded area is the required area.

Area $={\int }_0^1\left(y_{line}-y_{curve}\right)dx$

Area $={\int }_0^1\left(x+1-2^x\right)dx$

Using $\int x^{n}dx=\frac{x^{n+1}}{n+1} \& \int a^{x}dx=\frac{a^x}{{\log }_{e}a},$ we get

Area $={\left[\frac{x^2}{2}+x-\frac{2^x}{{\log }_{e}2}\right]}_0^1$

Area $=\left(\frac{1}{2}+1-\frac{2}{{\log }_{e}2}\right)- \left(0+0-\frac{1}{{\log }_{e}2}\right)$

Area $=\frac{3}{2}-\frac{1}{{\log }_{e}2}$ sq units.