Question

The are bounded by the curve $y=4x-x^2$ and the $x$-axis, is

• A. $\frac{32}{3}$ sq.unit
• B. $\frac{31}{7}$ sq.unit
• C. $\frac{32}{7}$ sq.unit
• D. $\frac{34}{3}$ sq.unit

Answer 1

The correct option is A $\frac{32}{3}$ sq.unit

Equation for x-axis is $y=0$ .......$\left(1\right)$

and other curve is $y=4x-x^2$ ........$\left(2\right)$

substitute $y=0$ in $\left(2\right)$ we get

$4x-x^2=0$

$x(4-x)=0$

$\Rightarrow x=0$

or $x=4$

Therefore the curve cuts the x-axis at $(0,0)$ and $(4,0)$

The area in interval $[0,4]$ above the x-axis

Required area $={\int }_0^{4}ydx$

$={\int }_0^{4}4x-x^2$

$\Rightarrow {\int }_0^{4}4xdx-{\int }_0^4{x}^{2}dx$

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

$=\frac{4}{2}[16-0]-\frac{1}{3}[64-0]$

$=32-\frac{64}{3}=\frac{96-64}{3}=\frac{32}{3}$

Area$=\frac{32}{3}$ sq. units.