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Question

0aa2-x2dx=


A

πa2

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B

πa22

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C

πa23

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D

πa24

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Solution

The correct option is D

πa24


Explanation for the correct option:

Let I=0aa2-x2dx

Let us consider x=asinθ

dx=acosθ

I=0aa2-a2sin2θacosθdθ=0aa2(1-sin2θacosθdθ=0aa2cos2θacosθdθ=0aacosθacosθdθ=a20acos2θdθ

Now we know that, cos2θ=1+cos2θ2.

Substituting this value in the above integral we get,

I=a20a1+cos2θ2dθ=a2θ2+sin2θ40a+Ccos2θ=sin2θ=a2sin-1xa2+2sinθcosθ40a+Cx=asin(θ)θ=sin-1xaandsin2θ=2sinθcosθ=a2sin-1xa2+2xaa2-x2a40a+C=a2π4+0-0-0+C=a2π4+CI=πa24+C

Hence, option D is the correct answer.


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