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Integrals · Question 12

JEE Main 2026 (Online) 2nd April Morning ShiftMCQMedium📈 Topic in 71% papers+4−1
1

Let [.] denote the greatest integer function. Then the value of 03 ( ex+e-x[x]! )dx is :

  1. e2+e3- 1e2- 1e3
  2. 12 (e2+e3- 1e2- 1e3 )
  3. e2+e3- 12e2- 12e3
  4. 12 (e2+e3 )- 1e2- 1e3

Answer

(2) 12 (e2+e3- 1e2- 1e3 )

Solution

On [0,1): [x]!=0!=1; on [1,2): [x]!=1!=1; on [2,3): [x]!=2!=2

I= 02(ex+e-x) dx+ 12 23(ex+e-x) dx

= [ex-e-x ]02+ 12 [ex-e-x ]23

= (e2-e-2 )+ 12 (e3-e-3-e2+e-2 )

= 12e2+ 12e3- 12e-2- 12e-3

= 12 (e2+e3- 1e2- 1e3 )

Answer: (2)

JEE Main 2026 (Online) 2nd April Morning Shift

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2

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3

The value of 020 ( 4 x+ 4 x )dx is equal to:

  1. 15 2
  2. 25
  3. 15
  4. 25 2
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4

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  1. 31
  2. 37
  3. 39
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5

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  1. 332
  2. 53
  3. 272
  4. 413
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6

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  1. 1 3(26 - 3)
  2. 4 3(8 - 6)
  3. 1 3(26 + 3)
  4. 2 3(4 + 6)
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7

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8

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  1. -5
  2. -11
  3. -18
  4. -26
JEE Main 2026 (Online) 28th January Evening Shift
9

Let [.] denote the greatest integer function. Then - 2 2 ( 12(3 + [x])3 + [ x] + [ x] ) dx is equal to :

  1. 12 + 5
  2. 11 + 2
  3. 15 + 4
  4. 13 + 1
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10

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  1. 1 + 2
  2. -1 - 2 2
  3. -1 - 2
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11

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