Integrals

JEE Main 2025 (Online) 4th April Evening ShiftNumericalHard📈 Topic in 21% papers+4

Ques. 1JEE Main 2025, 4 April (Evening Shift)

If ( 1+x2+x )10 ( 1+x2-x )9 dx= 1m ( ( 1+x2+x )n (n 1+x2-x ) )+C where C is the constant of integration and m, n N, then m+n is equal to .

Answer

379

Solution

Let t= 1+x2+x. Then 1+x2-x= 1t

1+x2= 12 (t+ 1t ), x= 12 (t- 1t )

dtdx= x 1+x2+1= t 1+x2 dx= 12 (1+ 1t2 )dt

I= t10 t9 12 (1+ 1t2 )dt= 12 (t19+t17 )dt

I= t2040+ t1836+C

Right side: n 1+x2-x= 12 [(n-1)t+ n+1t ]

1mtn (n 1+x2-x )= (n-1)tn+1+(n+1)tn-12m

Comparing: n+1=20 n=19

n-12m= 140 182m= 140 m=360

Check: n+12m= 20720= 136 (matches)

m+n=360+19=379

Answer: 379

Keep practising — Integrals

Ques. 2JEE Main 2025, 3 April (Morning Shift)

Let f(x) = x3 3 - x2 dx. If 5f( 2) = -4, then f(1) is equal to

- 6 25
- 8 25
- 2 25
- 4 25

Ques. 3JEE Main 2025, 3 April (Morning Shift)

Let the domain of the function f(x) = 2 4 6 (3 + 4x - x2) be (a, b). If 0b-a [x2] dx = p - q - r, p, q, r N, (p, q, r) = 1, where [ ] is the greatest integer function, then p + q + r is equal to

Ques. 4JEE Main 2025, 3 April (Evening Shift)

The integral 0 8x dx4 2 x + 2 x is equal to

2 2
4 2
2
3 22

Ques. 5JEE Main 2025, 2 April (Morning Shift)

Let [.] denote the greatest integer function. If 0e3 [ 1ex-1 ] dx = - e 2, then 3 is equal to .

Ques. 6JEE Main 2025, 2 April (Evening Shift)

Let (a, b) be the point of intersection of the curve x2 = 2y and the straight line y - 2x - 6 = 0 in the second quadrant. Then the integral I = ab 9x21 + 5x dx is equal to :

Ques. 7JEE Main 2025, 2 April (Evening Shift)

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