Find the area of the region bounded by the lines $\displaystyle y=4 x+5, x+y=5$ and $\displaystyle 4 y=x+5$, using integration.

Correct Figure Vertices are $\displaystyle \mathrm{A}(0,5), \mathrm{B}(-1,1), \mathrm{C}(3,2)$, Required area $\displaystyle =\int_{-1}^{0}(4 x+5) d x+\int_{0}^{3}(5-x) d x-\int_{-1}^{3} \frac{x+5}{4} d x$ $\displaystyle =\left.\frac{(4 x+5)^{2}}{8}\right|_{-1} ^{0}+\left.\frac{(5-x)^{2}}{-2}\right|_{0} ^{3}-\left.\frac{(x+5)^{2}}{8}\right|_{-1} ^{3}$ $\displaystyle =3+\frac{21}{2}-6=\frac{\mathbf{1 5}}{\mathbf{2}}$