How do you find the area between f(x)=3(x^3-x) and g(x)=0?

Mar 1, 2017

0

Explanation:

Always draw a graph or sketch when finding an area so that you get an understanding of for the area in question:

$f \left(x\right) = 3 \left({x}^{3} - x\right)$

graph{ 3(x^3-x) [-2, 2, -3, 3]}

We can easily show that f(x) is an odd function:

$f \left(- x\right) = 3 \left({\left(- x\right)}^{3} - \left(- x\right)\right)$
$\text{ } = 3 \left(- {x}^{3} + x\right)$
$\text{ } = - 3 \left({x}^{3} - x\right)$
$\text{ } = - 3 f \left(x\right)$

The purpose of demonstrating this property is because now we can conclude that the area bound by the curve $f \left(x\right) = 3 \left({x}^{3} - x\right)$ and $g \left(x\right) \equiv 0$ to the right of $O y$ is identical to that to the left of $O y$. As one has positive contribution, and the other has negative contribution, then the net area is zero.

If however we wanted the positive contribution area only, then this would be given by:

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

Similarly, the negative contribution would be $- \frac{3}{4}$

You should interpret these result depending upon the nature of the initial question!