The value of α for which 4 α ∫ - 1 2 e - α x d x = 5 , is

The value of α for which 4α-12e-αxdx=5 , is
  1. loge2
  2. loge32
  3. loge2
  4. loge43

Solution

4α-10eαxdx+02e-αxdx=5

4αeαxα-10+e-αx-α02=5

4α1-e-αα-e-2α-1α=5

42-e-α-e-2α=5

Put e-α=t

4t2+4t-3=02t+32t-1=0

e-α=12α=ln2

Asked in: JEE Main 2020 (07 Jan Shift 2)

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