# What is the area between the graphs?

May 23, 2018

Ω=5/12m^2

#### Explanation:

Ω=int_0^1(root(3)(x)-x^2)dx=

${\int}_{0}^{1} \sqrt[3]{x} \mathrm{dx} - {\int}_{0}^{1} {x}^{2} \mathrm{dx} =$

${\int}_{0}^{1} {x}^{\frac{1}{3}} \mathrm{dx} - {\int}_{0}^{1} {x}^{2} \mathrm{dx} =$

${\left[\frac{3}{4} {x}^{\frac{4}{3}}\right]}_{0}^{1} - {\left[{x}^{3} / 3\right]}_{0}^{1}$

$\frac{3}{4} - \frac{1}{3} = \frac{5}{12} {m}^{2}$

May 23, 2018

$\frac{5}{12}$

#### Explanation:

the integral is the area between the blue and red curves from the green line $\left(x = 0\right)$ and the orange line $\left(x = 1\right)$:

the area is ${\int}_{0}^{1} \left({x}^{\frac{1}{3}} - {x}^{2}\right) \mathrm{dx}$ (subtract ${x}^{2}$ from ${x}^{\frac{1}{3}}$ because ${x}^{\frac{1}{3}}$ is always greater on $0 < x < 1$)

solving:
${\int}_{0}^{1} \left({x}^{\frac{1}{3}} - {x}^{2}\right) \mathrm{dx} = F \left(1\right) - F \left(0\right)$, where $F \left(x\right) = \frac{3}{4} {x}^{\frac{4}{3}} - \frac{1}{3} {x}^{3}$

$= \frac{3}{4} {\left(1\right)}^{\frac{4}{3}} - \frac{1}{3} {\left(1\right)}^{3} - \frac{3}{4} {\left(0\right)}^{\frac{4}{3}} + \frac{1}{3} {\left(0\right)}^{3}$
$= \frac{3}{4} - \frac{1}{3} = \frac{5}{12}$