First slide

Evaluation of definite integrals

Question

$\int_{\frac{π}{2}}^{\frac{3 π}{2}} [2 \sin x] d x =$

Difficult

A

$- π$
B

0
C

$- \frac{π}{2}$
D

$\frac{π}{2}$

Solution

$x = \frac{π}{2} + t \to I = \int_{2}^{π} [2 \cos t] d t$
$= \int_{0}^{\frac{π}{3}} 1 d t + \int_{\frac{π}{3}}^{\frac{π}{2}} 0 \cdot d t + \int_{\frac{π}{2}}^{\frac{2 π}{3}} (- 1) d t + \int_{\frac{2 π}{3}}^{π} (- 2) d t$

$= \frac{π}{3} - \frac{π}{6} - \frac{2 π}{3} = - \frac{π}{2}$

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