Question

The area, in square unit, 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 $\frac{17}{12}$

The required area = ${\int }_1^2(x^3-x^2)dy=\frac{x^4}{4}-\frac{x^3}{3}{]}_1^2=(4-\frac{8}{3})–(\frac{1}{4}-\frac{1}{3})=\frac{48-32-3+4}{2}=\frac{17}{12}$