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Application of Integrals

JEE Main 2025 (Online) 24th January Morning ShiftMCQMedium📈 Topic in 71% papers+4−1
Ques. 1JEE Main 2025, 24 January (Morning Shift)

The area of the region \(x,y):x2+4x+2 y |x+2| \ is equal to

  1. 7
  2. 24/5
  3. 20/3
  4. 5

Answer

(3) 20/3

Solution

Shift u=x+2: region u2-2 y |u|, symmetric about the y-axis; the curves meet at |u|=2.

[figure]

Area =2 02 (u-u2+2 )du=2 [2- 83+4 ]= 203

Answer: (3)

Keep practising — Application of Integrals

Ques. 2JEE Main 2025, 24 January (Evening Shift)

The area of the region enclosed by the curves y=ex, y=|ex-1| and y-axis is :

  1. 1+ e2
  2. e2
  3. 1- e2
  4. 2 e2-1
Ques. 3JEE Main 2025, 23 January (Morning Shift)

If the area of the larger portion bounded between the curves x2+y2=25 and y=|x-1| is 14(b +c), b,c , then b+c is equal to

Ques. 4JEE Main 2025, 23 January (Evening Shift)

If the area of the region \(x,y):-1 x 1, 0 y a+e|x|-e-x, a>0 \ is e2+8e+1e, then the value of a is :

  1. 7
  2. 5
  3. 6
  4. 8
Ques. 5JEE Main 2025, 22 January (Morning Shift)

The area of the region, inside the circle (x-2 3)2+y2=12 and outside the parabola y2=2 3 x is :

  1. 3 -8
  2. 6 -8
  3. 3 +8
  4. 6 -16
Ques. 6JEE Main 2025, 22 January (Morning Shift)

Let f:R be a twice differentiable function such that f(x+y)=f(x)f(y) for all x,y . If f'(0)=4a and f satisfies f''(x)-3af'(x)-f(x)=0, a>0, then the area of the region R=\(x,y) 0 y f(ax), 0 x 2\ is :

  1. e2-1
  2. e4+1
  3. e2+1
  4. e4-1
Ques. 7JEE Main 2025, 22 January (Morning Shift)

Let the function, f(x)= cases-3ax2-2, & x<1\ a2+bx, & x 1 cases be differentiable for all x , where a>1, b . If the area of the region enclosed by y=f(x) and the line y=-20 is + 3, , Z, then the value of + is .

Ques. 8JEE Main 2025, 22 January (Evening Shift)

The area of the region enclosed by the curves y=x2-4x+4 and y2=16-8x is :

  1. 83
  2. 5
  3. 8
  4. 43

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