Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

Wat is de gemiddelde waarde ( = middelwaarde ) van de functie f (x) = 4 − x² over het interval [ − 2, 2 ] ?	A. \(\boldsymbol{\frac{16}{3}}\)
B. \(\boldsymbol{2}\)
C. \(\boldsymbol{\frac{8}{3}}\)
D. \(\boldsymbol{4}\)
E. \(\boldsymbol{\frac{4}{3}}\)
F. \(\boldsymbol{\frac{32}{3}}\)
G. \(\boldsymbol{\frac{128}{3}}\)

What is the average value of the function f (x) = 4 − x² over the interval [ − 2, 2 ] ?	A. $\boldsymbol{\frac{16}{3}}$
	B. $\boldsymbol{2}$
	C. $\boldsymbol{\frac{8}{3}}$
	D. $\boldsymbol{4}$
	E. $\boldsymbol{\frac{4}{3}}$
	F. $\boldsymbol{\frac{32}{3}}$
	G. $\boldsymbol{\frac{128}{3}}$

$\\\frac{1}{2-(-2)}\int_{-2}^{2}(4-x^2)\;dx=\frac14\left[4x-\frac{x^3}{3}\right]_{-2}^2=\frac14\left[\left(4.2-\frac{2^3}{3}\right)-\left(4(-2)-\frac{(-2)^3}{3}\right)\right]\\=\frac14\left[8-\frac83+8-\frac83\right]=\frac14\left[16-\frac{16}{3}\right]=4-\frac43=\frac{12-4}{3}=\frac83$
$\\\frac{1}{2-(-2)}\int_{-2}^{2}(4-x^2)\;dx=\frac14\left[4x-\frac{x^3}{3}\right]_{-2}^2=\frac14\left[\left(4.2-\frac{2^3}{3}\right)-\left(4(-2)-\frac{(-2)^3}{3}\right)\right]\\=\frac14\left[8-\frac83+8-\frac83\right]=\frac14\left[16-\frac{16}{3}\right]=4-\frac43=\frac{12-4}{3}=\frac83$