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 x-axis, so 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 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 above graph 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}


Which shows that the area under the curve.

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