Answer to Question #151569 in Calculus for Joshua

Question

Find the area bounded by the curve y = x^3-12 x and the x-axis. Sketch the curve indicating the area bounded.

Answer 1

Solution. Find the intersection points of the lines

$$y=x^3-12x$$ $$y=0$$

Therefore get $x^3-12x=0$

$$x(x^2-12)=0$$

The roots of the equation are

$$x_1=0$$ $$x_2=\sqrt{12}=2\sqrt{3}$$ $$x_3=-\sqrt{12}=-2\sqrt{3}$$

Sketch the curve indicating the area bounded.

Find the area bounded by the curve y = x^3-12 x and the x-axis. (considering the symmetry of the figure)

$$A=2\int_{-2\sqrt{3}}^0(x^3-12x-0)dx=\int_{-2\sqrt{3}}^0(2x^3-24x)dx=$$

$$=(\frac{x^4}{2}-12x^2)|_{-2\sqrt{3}}^0=0-72+144=72$$

Answer. 72.