Consider the function $f(x)=-x^2+4$. Determine the area of the region enclosed by the graph of $f(x)$, the $x$-axis and the lines $x=0$ and $x=3$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$3$
Antwoord 2 correct
Fout
Antwoord 3 optie
$5$
Antwoord 3 correct
Fout
Antwoord 4 optie
$5\frac{1}{3}$
Antwoord 4 correct
Fout
Antwoord 1 optie
$7\frac{2}{3}$
Antwoord 1 feedback
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.
$$\begin{align}
O_1 &= \int_{0}^{2} f(x)dx =[-\tfrac{1}{3}x^3+4x]_{0}^{2}=5\frac{1}{3}\\
O_2 &= -\int_{2}^3 f(x)dx =-[-\tfrac{1}{3}x^3+4x]_{2}^{3}=2\frac{1}{3}
\end{align}$$
Hence, $O_1+O_2= (5\frac{1}{3}) + (2\frac{1}{3}) = 7\frac{2}{3}$.
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: You first have to find an antiderivative, before plugging in the values $x=0$ and $x=3$.
See Integral.
See Integral.
Antwoord 4 feedback
Wrong: Calculate the integral over the entire interval $[0,3]$, not just over the subinterval $[0,2]$.
See Example (film).
See Example (film).