\section*{Exercise 3 - V11223}

Discuss the function

\[
f(x)= \begin{cases}\frac{|x|}{x} & \text { for }|x|>1 \\ x^{3} & \text { for }|x| \leq 1\end{cases}
\]

Calculate the area of the region bounded by the graph of the function, the x-axis, and the lines \(x=-2\) and \(x=2\)!
}

1) We are discussing the function. It holds that

\[
f(x)= \begin{cases}-1 & \text { for } x<0 \\ x^3 & \text { for } 0 \leq x \leq 1 \\ 1 & \text { for } x>1\end{cases}
\]

for the function. Thus, the function \(f\) is \(x^3\) for \(0 \leq x \leq 1\). At the point \(x=1\), the transition is continuous but not differentiable. Furthermore, the function \(f\) is constant with a function value of 1 for \(x>1\).

The function \(f\) is point-symmetric with respect to the coordinate origin, since

\[
-f(-x)=f(x)
\]

holds for all \(x \in \mathbb{R}\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1722.jpg?height=331&width=734&top_left_y=1405&top_left_x=661)

2) Due to the point symmetry, the considered area can be calculated as follows:

\[
A_{\text {total}}=2 \cdot \int_{0}^{2} f(x) \, \mathrm{d} x=2 \cdot\left\{\int_{0}^{1} x^{3} \, \mathrm{d} x+\int_{1}^{2} 1 \, \mathrm{d} x\right\}
\]

Therefore, it holds that

\[
A_{\text {total}}=2 \cdot\left\{\left[\frac{x^{4}}{4}\right]_{0}^{1}+1\right\}=2 \cdot \frac{5}{4}=\frac{5}{2}
\]