Mathematics - Area Under Curves Question with Solution | TestHub

From the previous question, the total area of region $R$ is $A=\frac{9}{2}$. The area $A_1$ for $0\le x\le 1$ is given by:$A_1={\int }_0^1(3x-x^2)dx={\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]}_0^1=\frac{3(1{)}^2}{2}-\frac{1^3}{3}=\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6}$ The area $A_2$ for $1\le x\le 3$ is $A-A_1$:$A_2=\frac{9}{2}-\frac{7}{6}=\frac{27-7}{6}=\frac{20}{6}=\frac{10}{3}$ The required ratio is $\frac{A_2}{A_1}$:$\frac{A_2}{A_1}=\frac{10/3}{7/6}=\frac{10}{3}\times \frac{6}{7}=\frac{20}{7}$ The ratio is $\frac{20}{7}\approx 2.86$ (correct to two decimal places).