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Integrals · Question 46

JEE Main 2025 (Online) 2nd April Evening ShiftMCQMedium📈 Topic in 3% papers+4−1
1

4 01 ( 1 3 + x2 + 1 + x2 )dx - 3 e( 3) is equal to :

  1. 2 - 2 - e(1 + 2)
  2. 2 + 2 + e(1 + 2)
  3. 2 + 2 - e(1 + 2)
  4. 2 - 2 + e(1 + 2)

Answer

(1) 2 - 2 - e(1 + 2)

Solution

Rationalise: 1 3 + x2 + 1 + x2 = 3 + x2 - 1 + x2(3 + x2) - (1 + x2) = 3 + x2 - 1 + x22

4 01 ( ) dx = 2 01 3 + x2 dx - 2 01 1 + x2 dx

Using x2 + a2 dx = x2 x2 + a2 + a22 e |x + x2 + a2 | + C:

01 3 + x2 dx = 12(2) + 32 e 3 - 32 e 3 = 1 + 34 e 3

01 1 + x2 dx = 22 + 12 e(1 + 2)

4 01 ( ) dx = 2 + 32 e 3 - 2 - e(1 + 2)

3 e 3 = 32 e 3

Required value = 2 - 2 - e(1 + 2)

Answer: (1)

JEE Main 2025 (Online) 2nd April Evening Shift

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2

The integral 80 0 4 ( + 9 + 16 2 )d is equal to :

  1. 3 4
  2. 4 3
  3. 6 43
  4. 2 3
JEE Main 2025 (Online) 29th January Morning Shift
3

If - 2 2 96x2 2 x(1 + ex) dx = ( 2 + ), , Z, then ( + )2 equals

  1. 196
  2. 100
  3. 64
  4. 144
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4

Let f: R R be a twice differentiable function such that f(2) = 1. If F(x) = x f(x) for all x R, 02 x F'(x) dx = 6 and 02 x2 F''(x) dx = 40, then F'(2) + 02 F(x) dx is equal to:

  1. 13
  2. 11
  3. 9
  4. 15
JEE Main 2025 (Online) 28th January Evening Shift
5

If f(x) = 1x1/4 (1 + x1/4 )dx, f(0) = -6, then f(1) is equal to :

  1. 4 ( e 2 - 2 )
  2. e2 2 + 2
  3. 2 - e2
  4. 4 ( e 2 + 2 )
JEE Main 2025 (Online) 28th January Evening Shift
6

Let f be a real valued continuous function defined on the positive real axis such that g(x) = 0x t f(t) dt. If g (x3 ) = x6 + x7, then value of r=115 f (r3 ) is :

  1. 270
  2. 340
  3. 310
  4. 320
JEE Main 2025 (Online) 28th January Evening Shift

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