If f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x , x ≥ 0 , and f 0 = 0 , then the value of f ( 1 )…

If fx=∫5x8+7x6x2+1+2x72dx, x≥0, and f0=0, then the value of f(1) is
  1. -14
  2. 12
  3. 14
  4. -12

Solution

We have, fx=∫5x8+7x6x2+1+2x72 dx

=∫5x8+7x6x141x5+1x7+22dx

=∫5x6+7x81x5+1x7+22dx

Put, 1x5+1x7+2=t

⇒-5x6-7x8dx=dt

∴fx=-∫dtt2=1t+c, where c is the constant of integration.

=x7x2+1+2x7+c

Now, f0=0⇒ c=0

∴  f1=112+2+2.17=14

Asked in: JEE Main 2019 (09 Jan Shift 2)

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