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Application of Integrals

JEE Main 2025 (Online) 22nd January Morning ShiftMCQMedium📈 Topic in 21% papers+4−1
Ques. 1JEE 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

Answer

(4) 6 -16

Solution

The circle and the parabola meet at (0,0) and (2 3, 2 3); the shaded part is the required area.

[figure]

By symmetry, area =2 02 3 [ 12-(x-2 3)2- 2 3 x ]dx

Circle part = quarter circle =3 ; parabola part = 23(2 3)2=8

area =2(3 -8)=6 -16

Answer: (4)

Keep practising — Application of Integrals

Ques. 2JEE 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. 3JEE 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. 4JEE 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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