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Question

101x2dx=

A
1π4
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B
1π3
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C
π3
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D
π4
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Solution

The correct option is D π4

101x2dx
1x2dx=x2a2x2+a22sin1(xa)+c
101x2dx=(1212x2+a22sin1(xa)+c)10
=(12(0)+12sin1(1)+c)(0+0+c)
=12π2
=π4
101x2dx=π4


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