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Question

Evaluate the integral 10ππ|sinx|dx

A
20
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B
8
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C
10
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D
18
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Solution

The correct option is D 18
For a periodic function f(x) with period a,

namaf(x)dx=(nm)a0f(x)dx

Here , f(x)=|sinx| whose period is π.

Hence, 10ππ|sinx|dx

=9π0|sinx|dx

=9π0sinxdx Since, sinx0x[0,π]

=9[cosx]π0

=18

Hence, answer is option-(D).

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