lf the value of the integral ∫01/2 x21−x23/2dx is k6, then k is equal to

lf the value of the integral 01/2x21x23/2dx is k6, then k is equal to

  1. A

    23+π

  2. B

    23-π

  3. C

    32+π

  4. D

    32-π

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    Solution:

    We have, 

    01/2x21x23/2dx=k6

    0π/6sin2θcos3θcosθdθ=k6, where x=sinθ

     0π/6tan2θdθ=k6 0π/6sec2θ1dθ=k6 [tanθθ]0π/6=k613π6=k6k=23π.

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