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Question

Evaluate aaaxa+xdx

A
πa
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B
πa2
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C
2a
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D
π2a
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Solution

The correct option is A πa
aaaa2x2dxaaxa2x2dx
=2aa01a2x2dx0[xa2x2is odd function]
=2a sin1xa|a0=2a[sin1(1)sin1(0)]=πa

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