# The area of the region bounded by the curve y=x^2 and y=4x-x^2 is ?

May 24, 2018

$\textcolor{b l u e}{A = {\int}_{0}^{2} \left(4 x - {x}^{2}\right) - \left({x}^{2}\right) \cdot \mathrm{dx} = \frac{8}{3} = 2.667 {\left(u n i t e\right)}^{2}}$

#### Explanation:

Firstly lets find the limits of the Area

since $y = {x}^{2}$ and $y = 4 x - {x}^{2}$

${x}^{2} = 4 x - {x}^{2}$

$2 {x}^{2} - 4 x = 0 \Rightarrow 2 x \cdot \left(x - 2\right) = 0$

$2 x = 0 \Rightarrow x = 0$

since $x = 0 \Rightarrow y = {x}^{2} \Rightarrow y = 0$

$A = \left(0 , 0\right)$

$x - 2 = 0 \Rightarrow x = 2$

since $x = 2 \Rightarrow y = {x}^{2} \Rightarrow y = 4$

$B = \left(2 , 4\right)$

if we take dx slice (that mean the Area revolving about x-axis)

The interval of the integral become

$x \in \left[0 , 2\right]$

now let set up the integral of Area

color(red)[A=int_a^by_2-y_1*dx

$A = {\int}_{0}^{2} \left(4 x - {x}^{2}\right) - \left({x}^{2}\right) \cdot \mathrm{dx}$

$A = {\int}_{0}^{2} 4 x - 2 {x}^{2} \cdot \mathrm{dx} = {\left[2 {x}^{2} - \frac{2}{3} {x}^{3}\right]}_{0}^{2}$

$\left[8 - \frac{16}{3}\right] = \frac{24 - 16}{3} = \frac{8}{3} = 2.667 {\left(u n i t e\right)}^{2}$

show the graph of the area below: