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Question: Answered & Verified by Expert
If the integral I=2x24+x2dx= 2x-fx+c, where f2=π, then the minimum value of y=fx x-2,2 is (where, c is the constant of integration)
MathematicsIndefinite IntegrationJEE Main
Options:
  • A 0
  • B -π
  • C 2π
  • D -4π
Solution:
2517 Upvotes Verified Answer
The correct answer is: -π
I=2x2+4-4x2+4dx
=21-4x2+4dx
=2x-42tan-1x2+c
=2x-4tan-1x2+c
fx=4tan-1x2
So, minfx=4tan-1-1=-π

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