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Application of Integrals

JEE Main 2024 (Online) 30th January Morning ShiftMCQEasy📈 Topic in 72% papers+4−1
Ques. 1JEE Main 2024, 30 January (Morning Shift)

The area (in square units) of the region bounded by the parabola y2=4(x-2) and the line y=2x-8, is :

  1. 7
  2. 8
  3. 9
  4. 6

Answer

(3) 9

Solution

Line: x= y+82; parabola: x= y24+2

y+82= y24+2 y2-2y-8=0 y=-2, 4

[figure]

for -2<y<4 the line is right of the parabola:

Area = -24 [ y+82- ( y24+2 ) ]dy= [ y24+2y- y312 ]-24

= 203- (- 73 )=9

Answer: (3)

Keep practising — Application of Integrals

Ques. 2JEE Main 2024, 30 January (Evening Shift)

The area of the region enclosed by the parabola (y-2)2=x-1, the line x-2 y+4=0 and the positive coordinate axes is .

Ques. 3JEE Main 2024, 29 January (Morning Shift)

The area (in sq. units) of the part of the circle x2+y2=169 which is below the line 5x-y=13 is 2 - 652+ -1 ( 1213 ), where , are coprime numbers. Then + is equal to .

Ques. 4JEE Main 2024, 29 January (Evening Shift)

Let the area of the region \(x, y): 0 x 3, 0 y \x2+2, 2x+2 \ \ be A. Then 12 A is equal to .

Ques. 5JEE Main 2024, 27 January (Morning Shift)

Let the area of the region \(x, y): x-2y+4 0, x+2y2 0, x+4y2 8, y 0 \ be mn, where m and n are coprime numbers. Then m+n is equal to .

Ques. 6JEE Main 2024, 27 January (Evening Shift)

If the area of the region \(x,y):0 y \2x,6x-x2 \ \ is A, then 12A is equal to .

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