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 $$y = {x^3} – x$$.


area-bounded-curve-x-axis

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
\[A = \int\limits_{ – 1}^0 {ydx} + \int\limits_0^1 { – ydx} \]
\[\begin{gathered} A = \int\limits_{ – 1}^0 {\left( {{x^3} – x} \right)dx} – \int\limits_0^1 {\left( {{x^3} – x} \right)dx} \\ \Rightarrow A = \left| {\frac{{{x^4}}}{4} – \frac{{{x^2}}}{2}} \right|_{ – 1}^0 – \left| {\frac{{{x^4}}}{4} – \frac{{{x^2}}}{2}} \right|_0^1 = 0 – \left( {\frac{1}{4} – \frac{1}{2}} \right) – \left( {\frac{1}{4} – \frac{1}{2}} \right) + 0 \\ \Rightarrow A = – \frac{1}{4} + \frac{1}{2} – \frac{1}{4} + \frac{1}{2} = \frac{1}{2} \\ \end{gathered} \]
\[Area = \frac{1}{2}\]