# What is the area between the line y = x and the curve y = x^3?

Jan 4, 2017

Area $= \frac{1}{2} \setminus {\text{unit}}^{2}$

#### Explanation:

Graphs of the functions $y = {x}^{3}$ and $y = x$ are:
graph{(y-x^3)(y-x)=0 [-3.465, 3.464, -1.733, 1.73]}

Both $y = {x}^{3}$ and $y = x$ are odd functions so by symmetry we only need to consider the RHS and double it. We can see from the graph that the (positive) intersection points are $\left(0 , 0\right)$ and $\left(1 , 1\right)$ which we can verify algebraical:

${x}^{3} = x \implies {x}^{3} - x = 0 \implies x \left({x}^{2} - 1\right) = 0 \implies x - 0 , \pm 1$

The area bounded by the $x$-axis, $y$-axis and $y = x$ is that of a triangle with area:

${A}_{1} = \frac{1}{2} \left(1\right) \left(1\right) = \frac{1}{2}$

The area bounded by the $x$-axis, $y$-axis and $y = {x}^{3}$ is given by:

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

Then the total bounded area that we seek (LHS and RHS) is given by:

$A = 2 \left({A}_{1} - {A}_{2}\right)$
$\setminus \setminus \setminus = 2 \left(\frac{1}{2} - \frac{1}{4}\right)$
$\setminus \setminus \setminus = \frac{1}{2}$

Jan 4, 2017

$\frac{1}{2}$ unit${\text{s}}^{2}$

#### Explanation:

Start by finding the points of intersection by solving the system $\left\{\begin{matrix}y = {x}^{3} \\ y = x\end{matrix}\right.$

${x}^{3} = x$

${x}^{3} - x = 0$

$x \left({x}^{2} - 1\right) = 0$

$x \left(x + 1\right) \left(x - 1\right) = 0$

$x = 0 , 1 \mathmr{and} - 1$

So, we have to find the area in the interval 0 ≤ x ≤ 1 and in the interval -1 ≤ x ≤ 0. If you trace the graphs of the two functions, you will find that the area in -1 ≤ x ≤ 0 lies beneath the y-axis while the area in 0 ≤ x ≤1 will lie above the y-axis. We simply add the two areas up in this case.

Here is the graph:

As you can see, in the interval 0 ≤ x ≤ 1, the line $y = x$ lies above $y = {x}^{3}$ and in the interval -1 ≤ x ≤ 0, $y = {x}^{3}$ lies above $y = x$. Our integrals will therefore be ${\int}_{-} {1}^{0} \left({x}^{3} - x\right)$ and ${\int}_{0}^{1} \left(x - {x}^{3}\right)$

Evaluating ${\int}_{-} {1}^{0} \left({x}^{3} - x\right) \mathrm{dx}$

Integrate using the rule $\int \left({x}^{n}\right) \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right)$:

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

Evaluate using ${\int}_{a}^{b} F \left(x\right) \mathrm{dx} = f \left(b\right) - f \left(a\right)$, where $f ' \left(x\right) = F \left(x\right)$.

$= \frac{1}{4} {\left(0\right)}^{4} - \frac{1}{2} {\left(0\right)}^{2} - \left(\frac{1}{4} {\left(- 1\right)}^{4} - \frac{1}{2} {\left(- 1\right)}^{2}\right)$

$= - \frac{1}{4} + \frac{1}{2}$

$= \frac{1}{4}$

Evaluating ${\int}_{0}^{1} x - {x}^{3} \mathrm{dx}$

Use the same process as above:

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

$= \frac{1}{2} {\left(1\right)}^{2} - \frac{1}{4} {\left(1\right)}^{2} - \left(\frac{1}{2} {\left(0\right)}^{2} - \frac{1}{4} {\left(0\right)}^{4}\right)$

$= \frac{1}{4}$

Add the two areas to find the total area.

${A}_{\text{total}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Hence, the area between $y = {x}^{3}$ and $y = x$ is $\frac{1}{2}$ unit${\text{s}}^{2}$.

Hopefully this helps!