Exercise 1

Consider the function $f(x)=x^2-x-2$. Determine the area enclosed by the graph of $f(x)$, the $x$-axis and the lines $x=-2$ and $x=0$.
$3$
$\frac{2}{3}$
$4$
$\frac{11}{6}$
Consider the function $f(x)=x^2-x-2$. Determine the area enclosed by the graph of $f(x)$, the $x$-axis and the lines $x=-2$ and $x=0$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$\frac{2}{3}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$4$
Antwoord 3 correct
Fout
Antwoord 4 optie
$\frac{11}{6}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$3$
Antwoord 1 feedback
Correct: Note that the zeros of the function $f(x)$ are $x=-1$ and $x=2$. On the interval $[-2,0]$ the function changes signs once. It holds that $f(x) \geq 0$ for $-2 \leq x \leq -1$ and $f(x) \leq 0$ for $-1 \leq x \leq 0$. The desired area consists of two parts: $O_1$ and $O_2$:

\begin{align} O_1 &= \int_{-2}^{-1} f(x)dx=[\tfrac{1}{3}x^3-\tfrac{1}{2}x^2-2x]_{-2}^{-1}=\frac{11}{6}\\ O_2 &= -\int_{-1}^0 f(x)dx=-[\tfrac{1}{3}x^3-\tfrac{1}{2}x^2-2x]_{-1}^{0}=\frac{7}{6} \end{align}

Hence, $O_1+O_2=(\frac{11}{6})+(\frac{7}{6})=3$.

Go on.
Antwoord 2 feedback
Wrong: Note that $f(x)\leq 0$ for $-1 \leq x\leq 2$.

See Example (film).
Antwoord 3 feedback
Wrong: You have to find an antiderivative first, before plugging in the values $x=-2$ and $x=0$.

See Integral.
Antwoord 4 feedback
Wrong: You have to calculate the integral over the entire interval $[-2,0]$, not just on the interval $[-2,-1]$.

See Example (film).