Using integration, find the area of the region bounded by y =5 x +4, y =0, x =-1 and x =1.

Required Area $\displaystyle =\left|\int_{-1}^{\frac{-4}{5}} y\, d x\right|+\int_{\frac{-4}{5}}^{1} y\, d x$

Area $\displaystyle =-\int_{-1}^{\frac{-4}{5}}(5 x+4) d x+\int_{\frac{-4}{5}}^{1}(5 x+4) d x$

Area $\displaystyle =-\left(\dfrac{(5 x+4)^{2}}{10}\right)_{-1}^{\frac{-4}{5}}+\left(\dfrac{(5 x+4)^{2}}{10}\right)_{\frac{-4}{5}}^{1}=\dfrac{1}{10}+\dfrac{81}{10}=\dfrac{41}{5}$ or $\displaystyle \dfrac{41}{5}$ sq. units