Let a be a positive real number such that ∫ 0 a e x - x d x = 10 e - 9 where, [ x ] is the greatest…

Let a be a positive real number such that 0aex-xdx=10e-9 where, [x] is the greatest integer less than or equal to x. Then, a is equal to:
  1. 10-loge1+e
  2. 10+loge2
  3. 10+loge3
  4. 10+loge1+e

Solution

Given:

a>0

Let na<n+1, nW

Then,

a=a+a

Here, a=n
Now,

0aex-xdx=10e-9

0nexdx+naex-xdx=10e-9

n01exdx+naex-ndx=10e-9

ne-1+ea-n-1=10e-9

ne-n+ea-n-1=10e-9

ne-n-1+eae-n=10e-9

On comparing, we get

n=10

and,

-n-1+eae-n=-9

-10-1+ea-10=-9

ea-10=2

a-10logee=loge2

a=10+loge2

Asked in: JEE Main 2021 (20 Jul Shift 1)

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