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we de bovengrens vervangen door x = 3<br />
zodat het voorbeeld dan als volgt luidt:<br />
2<br />
� 3 2<br />
0<br />
1<br />
√ dx =<br />
9 − x2 =<br />
� t=π/6<br />
t=0<br />
� t=π/6<br />
2<br />
t=0<br />
1<br />
3 cos t dt<br />
3 cos t<br />
dt = � t � π/6<br />
0<br />
π<br />
=<br />
6
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