Exercise 2

Correct: Note that the zeros of the function $f(x)$ are $x=-2$ and $x=2$. Hence, on the interval $[0,3]$ the function changes signs once. It holds that $f(x) \geq 0$ for $0 \leq x \leq 2$ and $f(x) \leq 0$ for $2 \leq x \leq 3$. Hence, the desired area consists of two parts: $O_1$ and $O_2$:

$$\begin{align} O_1 &= \int_{0}^{2} f(x)dx =[-\tfrac{1}{3}x^3+4x]_{0}^{2}=5\frac{1}{3}\\ O_2 &= -\int_{2}^3 f(x)dx =-[-\tfrac{1}{3}x^3+4x]_{2}^{3}=2\frac{1}{3} \end{align}$$

Hence, $O_1+O_2= (5\frac{1}{3}) + (2\frac{1}{3}) = 7\frac{2}{3}$.

Go on.

Wrong: Note that $f(x)\leq 0$ for $x\geq 2$.

See Example (film).

Wrong: You first have to find an antiderivative, before plugging in the values $x=0$ and $x=3$.

See Integral.

Wrong: Calculate the integral over the entire interval $[0,3]$, not just over the subinterval $[0,2]$.

See Example (film).