12.59 Find the area of the figure bounded by the lines $y=x^{3}-4 x$ and $y=0$ for $x \geq 0$.

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12.59 First, let's construct the graph of the function $y=x^{3}-4 x$ for $x \geq 0$. We find the extrema:

$y^{\prime}=3 x^{2}-4$;

$3 x^{2}-4=0 ;$

$x= \pm \frac{2}{\sqrt{3}} \approx \pm 1.1$.

For

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-402.jpg?height=440&width=538&top_left_y=599&top_left_x=604)

$x \geq 0$ the function is decreasing

on $\left(0, \frac{2}{\sqrt{3}}\right)$, increasing

Fig. 12.11

on $\left(\frac{2}{\sqrt{3}}, \infty\right)$, and $x=\frac{2}{\sqrt{3}}$ is a point of minimum, with $y_{\min }=-\frac{16}{3 \sqrt{3}} \approx-3$. Therefore, we need to find the area of the figure $O A B$ (Fig. 12.11). Since the function $y=x^{3}-4 x$ takes negative values on $(0,2)$, we have $S=\left|\int_{0}^{2}\left(x^{3}-4 x\right) d x\right|$. We get

$\int_{0}^{2}\left(x^{3}-4 x\right) d x=\left.\left(\frac{x^{4}}{4}-4 \frac{x^{2}}{2}\right)\right|_{0} ^{2}=-4$,

from which we obtain $S=4$ (sq. units).

Answer: 4 sq. units.