Calculus problem #2206

To find our limits of integration, we must find where the curves intersect, so we set the two equations equal to each other. $$x^2=x^3-2x$$ $$x^3-2x-x^2=0$$ $$x(x+1)(x-2)=0$$ $$x=-1, \, 0,\, 2.$$ On the interval $(-1,0)$ , $x^3-2x>x^2$ , and on the interval $(0,2)$ , $x^2>x^3-2x$ , so the area is given by $$\int {-1}^{0} \left(x^3-2x-x^2\right) \, dx + \int 0^2 \left(x^2-x^3+2x\right) \, dx$$ $$=\left[\frac{1}{4}x^4 -x^2-\frac{1}{3}x^3\right] {-1}^0 + \left[\frac{1}{3}x^3 -\frac{1}{4}x^4+x^2 \right] 0^2$$ $$=\frac{37}{12} \approx \boxed{3.083}$$

First we need to know FROM where and TO where we are integrating. In other words, we need to know where those two curves meet:

$x^{2} = x^{3} - 2x \Rightarrow x = 0, x = -1, x = 2$

Then observe that the curve $y = x^{3} - 2x$ takes higher values than the curve $y = x^{2}$ for $x < 0$ . Then

$\left |\int_{-1}^{0} (x^{3} - 2x - x^{2})dx \right | + \left |\int_{0}^{2} (x^{2} - x^{3} + 2x) \right | = \frac{5}{12} + \frac{8}{3} = \frac{111}{36} \cong 3.08$

is the value of the area between those two curves.