Question

The area enclosed between the curve $x^2-3x-y=0$ and the line $y=x$, is

• A. $\frac{16}{3}$sq. units
• B. $\frac{8}{3}$ sq. units
• C. $\frac{22}{3}$ sq. units
• D. $\frac{32}{3}$ sq. units

Answer 1

The correct option is A $\frac{16}{3}$sq. units

$y=x$...(1)

$y=x^2-3x$...(2)

Equating (1) and (2)

$x=x^2-3x$

$\Rightarrow x=4$

Area$={\int }_0^4\left[\right(x^2-3x)+x]dx$

$={[\frac{x^3}{3}-\frac{3}{2}{x}^2+\frac{x^2}{2}]}_0^4$

$=\frac{64}{3}-16$

$=\frac{16}{3}$sq. units