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Integrals · Question 18

JEE Main 2026 (Online) 28th January Morning ShiftNumericalMedium📈 Topic in 71% papers+4
1

The value of r=120 ( | ( 0r x| x|dx ) | ) is .

Answer

210

Solution

Let Ir = 0r x| x|dx, r a positive integer.

| (r - x)| = | x|, by 0r f(x)dx = 0r f(r - x)dx:

Ir = 0r (r - x)| x|dx

Adding: 2Ir = r 0r | x|dx

| x| has period 1 0r | x|dx = r 01 x dx = r 2

2Ir = 2r2 Ir = r2

| Ir | = r2 = r

r=120 r = 20 212 = 210

Answer: 210

JEE Main 2026 (Online) 28th January Morning Shift

Keep practising — Integrals

2

Let f(x) = dxx2/3 + 2x1/2, be such that f(0) = -26 + 24 e(2). If f(1) = a + b e(3), where a, b Z, then a + b is equal to :

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

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
JEE Main 2026 (Online) 28th January Evening Shift
4

Let f(t) = ( 1 - ( e t)1 - ( e t) ) dt, t > 1. If f (e /2 ) = -e /2 and f (e /4 ) = e /4, then equals

  1. 1 + 2
  2. -1 - 2 2
  3. -1 - 2
  4. -1 + 2
JEE Main 2026 (Online) 24th January Morning Shift
5

If f(x) satisfies the relation f(x) = ex + 01 (y + xex) f(y) dy, then e + f(0) is equal to .

JEE Main 2026 (Online) 24th January Evening Shift
6

Let f(x) = (2 - x2 ) ex ( 1 + x )(1 - x)3/2 dx. If f(0) = 0, then f ( 12 ) is equal to:

  1. 2e - 1
  2. 2e + 1
  3. 3e - 1
  4. 3e + 1
JEE Main 2026 (Online) 23rd January Morning Shift
7

The value of the integral 24 5 24 dx1 + [3] 2x is :

  1. 3
  2. 18
  3. 6
  4. 12
JEE Main 2026 (Online) 23rd January Morning Shift
8

Let I(x) = 3 dx(4x + 6) ( 4x2 + 8x + 3 ) and I(0) = 34 + 20. If I ( 12 ) = a 2b + c, where a, b, c N, (a, b) = 1, then a + b + c is equal to :

  1. 30
  2. 29
  3. 28
  4. 31
JEE Main 2026 (Online) 23rd January Evening Shift
9

The number of elements in the set S = \x : x [0, 100] and 0x t2 (x - t) dt = x2 \ is

JEE Main 2026 (Online) 23rd January Evening Shift
10

The value of - 2 2 ( 1[x] + 4 ) dx, where [ ] denotes the greatest integer function, is

  1. 160(21 - 1)
  2. 160( - 7)
  3. 760( - 3)
  4. 760(3 - 1)
JEE Main 2026 (Online) 22nd January Morning Shift
11

If ( x) -112( x) -52 dx = - p1q1( x) 92 - p2q2( x) 52 - p3q3( x) 12 + p4q4( x) -32 + C, where pi and qi are positive integers with (pi, qi) = 1 for i = 1, 2, 3, 4 and C is the constant of integration, then 15p1p2p3p4q1q2q3q4 is equal to

JEE Main 2026 (Online) 22nd January Morning Shift

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