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Question

If k012+8x2dx=π16, then the value of k is ______.

A
12
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B
13
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C
14
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D
15
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Solution

The correct option is A 12
Apply Integral Substitution
x=222u=12(2+2u2)du=1212+2u2du=121211+u2du=1212tan1u=1212tan1222x=14tan12x14(tan12k0)=π16 k= 12

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