If the ratio in which the $x$-axis divide the area of the region bounded by the parabolas $y=4 x-x^{

Question

If the ratio in which the $x$-axis divide the area of the region bounded by the parabolas $y=4 x-x^{2}$ and $y=x^{2}-x$ ?

Professor PC Thomas Classes · Answer

Solving for points of intersection\ $O(0,0), A\left(\frac{5}{2}, \frac{15}{4}ight)$\ Area above the $x$-axis\ $=\left|\int_{0}^{5 / 2}\left(4 x-x^{2}ight) d x-\int_{1}^{5 / 2}\left(x^{2}-xight) d xight|$\ $=\frac{121}{24}=\Delta_{1}$\ Area below the $x$-axis\ $=\left|\int_{0}^{1}-\left(x-x^{2}ight) d xight|=\frac{1}{6}=\Delta_{2}$\ $	herefore \quad$ Ratio in which $x$ axis divides above and below is\ $\Delta_{1}: \Delta_{2}=\frac{121}{24}: \frac{1}{6}=121: 4$.\ \