# The Area Bounded by the Curve y=x^3-x and the x-axis

In this tutorial we shall find the area bounded by the curve $y = {x^3} – x$ and the x-axis.

Since $y = 0$ at the x-axis, for the points of intersection of the curve $y = {x^3} – x$ at x-axis, put $y = 0$. This implies that ${x^3} – x = 0$
$\begin{gathered} \Rightarrow x\left( {{x^2} – 1} \right) = 0 \\ \Rightarrow x\left( {x – 1} \right)\left( {x + 1} \right) = 0 \\ \Rightarrow x = – 1,\,\,\,x = 0,\,\,\,x = 1 \\ \end{gathered}$

The curve cuts the x-axis only at $x = – 1,\,\,\,x = 0,\,\,\,x = 1$, as shown by the graph of the given function $y = {x^3} – x$. It is also clear from the graph above that $y \geqslant 0$ for $– 1 \leqslant x \leqslant 0$ and $y \leqslant 0$ for $0 \leqslant x \leqslant 1$, so the required area is split into two regions and is given by
$A = \int\limits_{ – 1}^0 {ydx} + \int\limits_0^1 { – ydx}$
$\begin{gathered} A = \int\limits_{ – 1}^0 {\left( {{x^3} – x} \right)dx} – \int\limits_0^1 {\left( {{x^3} – x} \right)dx} \\ \Rightarrow A = \left| {\frac{{{x^4}}}{4} – \frac{{{x^2}}}{2}} \right|_{ – 1}^0 – \left| {\frac{{{x^4}}}{4} – \frac{{{x^2}}}{2}} \right|_0^1 = 0 – \left( {\frac{1}{4} – \frac{1}{2}} \right) – \left( {\frac{1}{4} – \frac{1}{2}} \right) + 0 \\ \Rightarrow A = – \frac{1}{4} + \frac{1}{2} – \frac{1}{4} + \frac{1}{2} = \frac{1}{2} \\ \end{gathered}$
$Area = \frac{1}{2}$