Exercise 1 | Wiskunde op Tilburg University

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).