Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De gemiddelde waarde van tan 2x over het interval \(\boldsymbol{[\,0,\frac{\pi}{8}]}\) bedraagt	A. \(\frac{1}{2}\)
B. \(\frac{\pi}{4}\)
C. \(\frac{8}{\pi}\)
D. \(\frac{4}{\pi}.\ln 2\)
E. \(\frac{2}{\pi}.\ln 2\)

What is the average value of the function f (x) = tan 2x over the interval $\boldsymbol{[\,0,\frac{\pi}{8}]}$ ?	A. $\frac{1}{2}$
	B. $\frac{\pi}{4}$
	C. $\frac{8}{\pi}$
	D. $\frac{4}{\pi}.\ln 2$
	E. $\frac{2}{\pi}.\ln 2$

$\\\frac{1}{\frac{\pi}{8}-0}\int_{0}^{\frac{\pi}{8}}\tan 2x\,dx=\frac{8}{\pi}\int_{0}^{\frac{\pi}{8}}\frac{\sin2x}{\cos2x}\,dx=\frac{8}{\pi}\left(-\frac12\right)\int_{0}^{\frac{\pi}{8}}\frac{-2\sin2x}{\cos2x}\,dx\\=-\frac{4}{\pi}\left[\ln\cos2x\right]_0^{\frac{\pi}{8}}=-\frac{4}{\pi}\left[\ln\cos\frac{\pi}{4}-\ln\cos0\right]_0^{\frac{\pi}{8}}=-\frac{4}{\pi}\left[\ln\frac{\sqrt2}{2}-0\right]\\=-\frac{4}{\pi}\left[\ln\frac{1}{\sqrt2}\right]=-\frac{4}{\pi}(\ln1-\ln\sqrt2)=-\frac{4}{\pi}(-\frac12\ln2)=\frac{2}{\pi}\ln2$