First slide

Evaluation of definite integrals

Question

If $\int_{- 1}^{3 / 2} | x \sin π x | d x = k / π^{2},$ then the value of $k$ is

Difficult

A

$3 π + 1$
B

$2 π + 1$
C

1
D

4

Solution

$\begin{array}{l} I = \int_{- 1}^{1} | x \sin π x d x | + \int_{1}^{3 / 2} | x \sin π x | d x \\ = 2 \int_{0}^{1} x \sin π x d x - \int_{1}^{3 / 2} x \sin π x d x \\ = 2 {[\frac{- x \cos π x}{π} + \frac{\sin π x}{π^{2}}]}_{0}^{1} - [\frac{- x \cos (πx)}{π} + \frac{\sin (πx)}{π^{2}}] \\ = 2 [\frac{1}{π}] - [- \frac{1}{π^{2}} - \frac{1}{π}] = \frac{3 π + 1}{π^{2}}, so k = 3 π + 1 . \end{array}$

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