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Application of Integrals · Question 1

JEE Advanced 2026 Paper 2NumericalMedium📈 Topic in 33% papers+4
1

Let be the area of the region enclosed between the curves C1, C2, and the lines x= 1 and x= 4. Then the value of - 1 e ( -2 e- 2 ) is .

Answer

2.4 to 2.6

Solution

1=0, 2= 2, 3=2 , 4= 5 2.

ddx [e-x(1- x) ]=e-x( x+ x-1), so take F(x)=e-x(1- x).

x+ x-1 is >0 on (0, 2 ), <0 on ( 2,2 ), >0 on (2 , 5 2 ).

F(0)=0, F ( 2 )=e- /2, F(2 )=0, F ( 5 2 )=e-5 /2.

=e- /2+e- /2+e-5 /2 -2e- /2=e-5 /2.

- 1 e (e-5 /2 )= 52=2.5.

Answer: 2.5

JEE Advanced 2026 Paper 2

Keep practising — Application of Integrals

2

If is the area of the common region that lies inside both the given ellipses, then the value of is .

JEE Advanced 2026 Paper 2
3

Let R denote the set of all real numbers. Then the area of the region \(x,y) R R:x>0, y> 1x, 5x-4y-1>0, 4x+4y-17<0 \ is

  1. 1716- e 4
  2. 338- e 4
  3. 578- e 4
  4. 172- e 4
JEE Advanced 2025 Paper 2
4

Let S denote the locus of the mid-points of those chords of the parabola y2=x, such that the area of the region enclosed between the parabola and the chord is 43. Let R denote the region lying in the first quadrant, enclosed by the parabola y2=x, the curve S, and the lines x=1 and x=4. Then which of the following statements is (are) TRUE?

  1. (4, 3) S
  2. (5, 2) S
  3. Area of R is 143-2 3
  4. Area of R is 143- 3
JEE Advanced 2025 Paper 2
5

Let S= \(x,y) R R: x 0, y 0, y2 4x, y2 12-2x and 3y+ 8 x 5 8 \. If the area of the region S is 2, then is equal to

  1. 172
  2. 173
  3. 174
  4. 175
JEE Advanced 2024 Paper 2

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