First slide

Evaluation of definite integrals

Question

$Evaluate \int_{- π}^{π} \frac{x \sin x d x}{e^{x} + 1}$

Moderate

A

$π$
B

2 $π$
C

3 $π$
D

4 $π$

Solution

$Let I = \int_{- π}^{π} \frac{x \sin x d x}{e^{x} + 1} - - - - - (1)$
$Using property IV, we replace x by 0 - x or - x$
$∴ I = \int_{- π}^{π} \frac{(- x) \sin (- x) d x}{e^{- x} + 1} = \int_{- π}^{π} \frac{e^{x} x \sin x d x}{e^{x} + 1} - - - - - (2)$
$Adding equations (1) and (2), we get 2 I = \int_{- π}^{π} \frac{(e^{x} + 1) x \sin x d x}{(e^{x} + 1)} 2 I = \int_{- π}^{π} x \sin x d x = 2 \int_{0}^{π} x \sin x d x$
$I = \int_{0}^{π} x \sin x d x$
$\begin{array}{r} I = \int_{0}^{π} (π - x) \sin (π - x) d x \end{array} I = \int_{0}^{π} π \sin x d x - I 2 I = π {(- \cos x)}_{0}^{π} 2 I = 2 π$

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