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Application of Integrals

JEE Main 2024 (Online) 30th January Evening ShiftNumericalMedium📈 Topic in 72% papers+4
Ques. 1JEE 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 .

Answer

5

Solution

x=2y-4 in (y-2)2=x-1 (y-3)2=0 the line touches the parabola at (2, 3)

[figure]

Area (horizontal strips) = 03 [(y-2)2+1 ]dy- 23(2y-4) dy

= [ (y-2)33+y ]03- [y2-4y ]23

= ( 13+3+ 83 )-(-3+4)=6-1=5

Answer: 5

Keep practising — Application of Integrals

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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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