Question

The area in square units bounded by the curves $y=x^3,y=x^2$ and the ordinates $x=1,x=2$ is

• A. $\frac{17}{12}$
• B. $\frac{12}{13}$
• C. $\frac{2}{7}$
• D. $\frac{7}{2}$

Answer 1

The correct option is A $\frac{17}{12}$

${\int }_1^2\left(x^3{-x}^2\right)dx={[\frac{x^4}{4}-\frac{x^3}{3}]}_1^2=[4-\frac{8}{3}]-[\frac{1}{4}-\frac{1}{3}]=\left[\frac{4}{3}\right]-[-\frac{1}{12}]=\frac{4}{3}+\frac{1}{12}=\frac{17}{12}$