First slide

Evaluation of definite integrals

Question

$\int_{0}^{1} | sin 2 π x | d x =$

Moderate

A

$2 π$
B

$\frac{2}{π}$
C

$\frac{π}{2}$
D

$π$

Solution

$\int_{0}^{1} | S i n 2 π x | d x = \int_{0}^{1 / 2} | S i n 2 π x | d x + \int_{1 / 2}^{1} | S i n 2 π x | d x$
                          $= \int_{0}^{1 / 2} S i n 2 π x d x + \int_{1 / 2}^{1} - S i n 2 π x d x$

                          $= {(\frac{- C o s 2 π x}{2 π})}_{0}^{\frac{1}{2}} + {(\frac{C o s 2 π x}{2 π})}_{\frac{1}{2}}^{1}$

                          $= \frac{4}{2 π} = \frac{2}{π}$

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