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Integrals · Question 29

JEE Main 2026 (Online) 22nd January Evening ShiftNumericalHard📈 Topic in 71% papers+4
1

Let [.] be the greatest integer function. If = 064 (x1/3 - [x1/3] )dx, then 1 0 ( 2 6 + 6 )d is equal to .

Answer

36

Solution

064 x1/3dx = 34 [x4/3 ]064 = 34 256 = 192

[x1/3] = k for k3 x < (k+1)3:

064 [x1/3]dx = 0(1) + 1(8 - 1) + 2(27 - 8) + 3(64 - 27)

= 7 + 38 + 111 = 156

= 192 - 156 = 36

Let f( ) = 2 6 + 6 ; f has period , so 036 f = 36 0 f

0 f = 0 /2 2 + 2 6 + 6 d (putting + 2 in the second half)

6 + 6 = 1 - 3 2 2 = 1 - 34 2 2

0 /2 d 1 - 34 2 2 = 0 /2 d 2 + 14 2 ( = 2 )

= 0 /2 4 2 d 4 + 2 = [2 -1 2 ]0 /2 =

1 036 f( ) d = 1 36 = 36

Answer: 36

JEE Main 2026 (Online) 22nd January Evening Shift

Keep practising — Integrals

2

The value of - /6 /6 ( + 4x111 - (|x| + /6) )dx is equal to:

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  3. 2
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3

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4

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5

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6

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  1. 2 + 3
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7

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  1. 12
  2. 9
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JEE Main 2025 (Online) 8th April Evening Shift
8

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  3. 3( +2)
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9

If ( 1x + 1x3 ) ( [23]3x-24 + x-26 ) dx = - 3( + 1) (3x + x ) + 1 + C, x > 0, ( , , Z), where C is the constant of integration, then + + is equal to .

JEE Main 2025 (Online) 7th April Evening Shift
10

The value of -11 (1 + |x| - x ) ex + ( |x| - x ) e-xex + e-x dx is equal to

  1. 1 + 2 23
  2. 1 - 2 23
  3. 2 + 2 23
  4. 3 - 2 23
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11

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JEE Main 2025 (Online) 4th April Evening Shift

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