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Application of Integrals · Question 9

JEE Main 2026 (Online) 2nd April Evening ShiftNumericalHard📈 Topic in 16% papers+4
1

If the area of the region bounded by 16x2-9y2=144 and 8x-3y=24 is A, then 3 (A+6 e(3) ) is equal to .

Answer

24

Solution

Hyperbola: x29- y216=1; line: y= 8x-243

Intersection: 16x2-(8x-24)2=144 x2-8x+15=0 x=3, 5

Points: (3, 0) and (5, 163 )

[figure]

For 3 x 5 the upper branch y= 43 x2-9 lies above the line.

A= 35 ( 43 x2-9- 8x-243 )dx

35 x2-9 dx= [ x2 x2-9- 92 e |x+ x2-9 | ]35

= (10- 92 e 9 )- (0- 92 e 3 )=10- 92 e 3

35 8x-243 dx= 13 [4x2-24x ]35= 13(-20+36)= 163

A= 43 (10- 92 e 3 )- 163=8-6 e 3

3 (A+6 e 3 )=3 8=24

Answer: 24

JEE Main 2026 (Online) 2nd April Evening Shift

Keep practising — Application of Integrals

2

The area of the region R = \(x, y) : xy 8, 1 y x2, x 0\ is

  1. 23(20 e(2) + 9)
  2. 13(40 e(2) + 27)
  3. 13(49 e(2) - 15)
  4. 23(24 e(2) - 7)
JEE Main 2026 (Online) 28th January Morning Shift
3

Let P1 : y = 4x2 and P2 : y = x2 + 27 be two parabolas. If the area of the bounded region enclosed between P1 and P2 is six times the area of the bounded region enclosed between the line y = x, > 0 and P1, then is equal to :

  1. 12
  2. 15
  3. 8
  4. 6
JEE Main 2026 (Online) 28th January Evening Shift
4

Let A1 be the bounded area enclosed by the curves y = x2 + 2, x + y = 8 and y-axis that lies in the first quadrant. Let A2 be the bounded area enclosed by the curves y = x2 + 2, y2 = x, x = 2, and y-axis that lies in the first quadrant. Then A1 - A2 is equal to

  1. 23(2 2 + 1)
  2. 23(3 2 + 1)
  3. 23( 2 + 1)
  4. 23(4 2 + 1)
JEE Main 2026 (Online) 24th January Morning Shift
5

Let f( ) denote the area of the region in the first quadrant bounded by x = 0, x = 1, y2 = x and y = | x - 5| - |1 - x| + x2. Then (f(0) + f(1)) is equal to

  1. 12
  2. 14
  3. 9
  4. 7
JEE Main 2026 (Online) 24th January Evening Shift
6

Let the area of the region bounded by the curve y = \ x, x\, lines x = 0, x = 3 2, and the x-axis be A. Then, A + A2 is equal to .

JEE Main 2026 (Online) 23rd January Morning Shift
7

The area of the region enclosed between the circles x2 + y2 = 4 and x2 + (y - 2)2 = 4 is:

  1. 23(4 - 3 3)
  2. 43(2 - 3)
  3. 43(2 - 3 3)
  4. 23(2 - 3 3)
JEE Main 2026 (Online) 23rd January Evening Shift
8

Let the line x = -1 divide the area of the region \(x, y) : 1 + x2 y 3 - x \ in the ratio m : n, (m, n) = 1. Then m + n is equal to

  1. 27
  2. 28
  3. 25
  4. 26
JEE Main 2026 (Online) 22nd January Morning Shift
9

The area of the region A = \(x, y) : 4x2 + y2 8 and y2 4x\ is:

  1. + 23
  2. 2 + 2
  3. + 4
  4. 2 + 13
JEE Main 2026 (Online) 22nd January Evening Shift
10

The area of the region, inside the ellipse x2 + 4y2 = 4 and outside the region bounded by the curves y = |x| - 1 and y = 1 - |x|, is :

  1. 2 - 1
  2. 3( - 1)
  3. 2( - 1)
  4. 2 - 12
JEE Main 2026 (Online) 21st January Morning Shift
11

If the area of the region \(x, y) : 1 - 2x y 4 - x2, x 0, y 0\ is , , N, ( , ) = 1, then the value of ( + ) is:

  1. 73
  2. 85
  3. 67
  4. 91
JEE Main 2026 (Online) 21st January Evening Shift

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