# How do you find the smaller area bounded by y=4x-x^3 and y=x^2-2x?

Feb 1, 2018

$5.33 {\text{ units}}^{2}$ ( 2 .d.p.)

#### Explanation:

From the graph we can see that the area we seek is in the interval $\left[0 , 2\right]$

For accuracy we need to find the points of intersection algebraically rather than trying to find them from the graph.

If:

$y = {x}^{2} - 2 x$ and $y = 4 x - {x}^{3}$

Then:

${x}^{2} - 2 x = 4 x - {x}^{3} \implies {x}^{3} + {x}^{2} - 6 x = 0$

Solving for $x$:

$x \left({x}^{2} - 6 x\right) = 0 \implies x \left(x + 3\right) \left(x - 2\right) = 0$

Points of intersection:

$\left(0 , 0\right) , \left(2 , 0\right) , \left(- 3 , 15\right)$

Area under $4 x - {x}^{3}$ in $\left[0 , 2\right]$

$\text{A} = {\int}_{0}^{2} \left(4 x - {x}^{3}\right) \mathrm{dx} = 2 {x}^{2} - \frac{1}{4} {x}^{4} = {\left[2 {x}^{2} - \frac{1}{4} {x}^{4}\right]}_{0}^{2}$

${\left[2 {x}^{2} - \frac{1}{4} {x}^{4}\right]}^{2} - {\left[2 {x}^{2} - \frac{1}{4} {x}^{4}\right]}_{0}$

Plugging in upper and lower bounds:

${\left[2 {\left(2\right)}^{2} - \frac{1}{4} {\left(2\right)}^{4}\right]}^{2} - {\left[2 {\left(0\right)}^{2} - \frac{1}{4} {\left(0\right)}^{4}\right]}_{0}$

${\left[8 - 4\right]}^{2} - {\left[0\right]}_{0}$

Area $= 4$

Area under ${x}^{2} - 2 x$ in $\left[0 , 2\right]$

$\text{A} = {\int}_{0}^{2} \left({x}^{2} - 2 x\right) \mathrm{dx} = \frac{1}{3} {x}^{3} - {x}^{2} = {\left[\frac{1}{3} {x}^{3} - {x}^{2}\right]}_{0}^{2}$

${\left[\frac{1}{3} {x}^{3} - {x}^{2}\right]}^{2} - {\left[\frac{1}{3} {x}^{3} - {x}^{2}\right]}_{0}$

Plugging in upper and lower bounds:

${\left[\frac{1}{3} {\left(2\right)}^{3} - {\left(2\right)}^{2}\right]}^{2} - {\left[\frac{1}{3} {\left(0\right)}^{3} - {\left(0\right)}^{2}\right]}_{0}$

${\left[\frac{8}{3} - 4\right]}^{2} - {\left[0\right]}_{0} = - \frac{4}{3}$

Area $= \frac{4}{3}$

Total area:

$4 + \frac{4}{3} = \frac{16}{3} = 5.33 {\text{ units}}^{2}$ ( 2 .d.p)