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Page 1
JEE Mains Previous Year Questions
(2021-2024): Applications of Integrals
2024
Q1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves ???? + 4 ?? = 16 and ?? + ?? = 6 is equal to :
(1) 28 - 30 log
?? ? 2
(2) 30 - 28 log
?? ? 2
(3) 30 - 32 log
?? ? 2
(4) 32 - 30 log
?? ? 2
Q2 - 2024 (01 Feb Shift 2)
Three points O ( 0 , 0 ) , P ( a , a
2
) , Q ( - b , b
2
) , a > 0 , b > 0, are on the parabola ?? = ?? 2
. Let ?? 1
be the area of the region bounded by the line PQ and the parabola, and ?? 2
be the area
of the triangle ???? ?? . If the minimum value of
?? 1
?? 2
is
?? ?? , g c d ? ( ?? , ?? ) = 1, then ?? + ?? is equal
to:
Q3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of ?? , for which area of the region bounded by
the parabolas 2 y
2
= kx and ky
2
= 2 ( y - x ) is maximum, is equal to :
Q4 - 2024 (27 Jan Shift 1)
Let the area of the region { ( ?? , ?? ) : ?? - 2 ?? + 4 = 0 , ?? + 2 ?? 2
= 0 , ?? + 4 ?? 2
= 8 , ?? = 0 } be
?? ?? ,
where ?? and n are coprime numbers. Then m + n is equal to
Q5 - 2024 (27 Jan Shift 2)
If the area of the region { ( ?? , ?? ) : 0 = ?? = m i n { 2 ?? , 6 ?? - ?? 2
} } is ?? , then 12 A is equal to.
Q6 - 2024 (29 Jan Shift 1)
Page 2
JEE Mains Previous Year Questions
(2021-2024): Applications of Integrals
2024
Q1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves ???? + 4 ?? = 16 and ?? + ?? = 6 is equal to :
(1) 28 - 30 log
?? ? 2
(2) 30 - 28 log
?? ? 2
(3) 30 - 32 log
?? ? 2
(4) 32 - 30 log
?? ? 2
Q2 - 2024 (01 Feb Shift 2)
Three points O ( 0 , 0 ) , P ( a , a
2
) , Q ( - b , b
2
) , a > 0 , b > 0, are on the parabola ?? = ?? 2
. Let ?? 1
be the area of the region bounded by the line PQ and the parabola, and ?? 2
be the area
of the triangle ???? ?? . If the minimum value of
?? 1
?? 2
is
?? ?? , g c d ? ( ?? , ?? ) = 1, then ?? + ?? is equal
to:
Q3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of ?? , for which area of the region bounded by
the parabolas 2 y
2
= kx and ky
2
= 2 ( y - x ) is maximum, is equal to :
Q4 - 2024 (27 Jan Shift 1)
Let the area of the region { ( ?? , ?? ) : ?? - 2 ?? + 4 = 0 , ?? + 2 ?? 2
= 0 , ?? + 4 ?? 2
= 8 , ?? = 0 } be
?? ?? ,
where ?? and n are coprime numbers. Then m + n is equal to
Q5 - 2024 (27 Jan Shift 2)
If the area of the region { ( ?? , ?? ) : 0 = ?? = m i n { 2 ?? , 6 ?? - ?? 2
} } is ?? , then 12 A is equal to.
Q6 - 2024 (29 Jan Shift 1)
The area (in sq. units) of the part of circle ?? 2
+ ?? 2
= 169 which is below the line 5 ?? - ?? =
13 is
????
2 ?? -
65
2
+
?? ?? sin
- 1
? (
12
13
) where ?? , ?? are coprime numbers. Then ?? + ?? is equal to
Q7 - 2024 (29 Jan Shift 2)
Let the area of the region { ( ?? , ?? ) : 0 = ?? = 3 , 0 = ?? = m i n { ?? 2
+ 2 , 2 ?? + 2 } } be ?? . Then 12 A
is equal to
Q8 - 2024 (30 Jan Shift 1)
The area (in square units) of the region bounded by the parabola y
2
= 4 ( x - 2 ) and the
line y = 2x - 8
(1) 8
(2) 9
(3) 6
(4) 7
Q9 - 2024 (30 Jan Shift 2)
Let ?? = ?? ( ?? ) be a curve lying in the first quadrant such that the area enclosed by the line
Y - y = Y
'
( x ) ( X - x ) and the co-ordinate axes, where ( ?? , ?? ) is any point on the curve, is
always
- ?? 2
2 Y
'
( x )
+ 1 , Y
'
( x ) ? 0. If Y ( 1 ) = 1, then 12Y ( 2 ) equals
Q10 - 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola ( ?? - 2 )
2
= ?? - 1, the line ?? - 2 ?? + 4 =
0 and the positive coordinate axes is
Q11 - 2024 (31 Jan Shift 1)
The area of the region
{ ( ?? , ?? ) : ?? 2
= 4 ?? , ?? < 4 ,
???? ( ?? - 1 ) ( ?? - 2 )
( ?? - 3 ) ( ?? - 4 )
> 0 , ?? ? 3 } is
(1)
16
3
(2)
64
3
(3)
8
3
(4)
32
3
Page 3
JEE Mains Previous Year Questions
(2021-2024): Applications of Integrals
2024
Q1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves ???? + 4 ?? = 16 and ?? + ?? = 6 is equal to :
(1) 28 - 30 log
?? ? 2
(2) 30 - 28 log
?? ? 2
(3) 30 - 32 log
?? ? 2
(4) 32 - 30 log
?? ? 2
Q2 - 2024 (01 Feb Shift 2)
Three points O ( 0 , 0 ) , P ( a , a
2
) , Q ( - b , b
2
) , a > 0 , b > 0, are on the parabola ?? = ?? 2
. Let ?? 1
be the area of the region bounded by the line PQ and the parabola, and ?? 2
be the area
of the triangle ???? ?? . If the minimum value of
?? 1
?? 2
is
?? ?? , g c d ? ( ?? , ?? ) = 1, then ?? + ?? is equal
to:
Q3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of ?? , for which area of the region bounded by
the parabolas 2 y
2
= kx and ky
2
= 2 ( y - x ) is maximum, is equal to :
Q4 - 2024 (27 Jan Shift 1)
Let the area of the region { ( ?? , ?? ) : ?? - 2 ?? + 4 = 0 , ?? + 2 ?? 2
= 0 , ?? + 4 ?? 2
= 8 , ?? = 0 } be
?? ?? ,
where ?? and n are coprime numbers. Then m + n is equal to
Q5 - 2024 (27 Jan Shift 2)
If the area of the region { ( ?? , ?? ) : 0 = ?? = m i n { 2 ?? , 6 ?? - ?? 2
} } is ?? , then 12 A is equal to.
Q6 - 2024 (29 Jan Shift 1)
The area (in sq. units) of the part of circle ?? 2
+ ?? 2
= 169 which is below the line 5 ?? - ?? =
13 is
????
2 ?? -
65
2
+
?? ?? sin
- 1
? (
12
13
) where ?? , ?? are coprime numbers. Then ?? + ?? is equal to
Q7 - 2024 (29 Jan Shift 2)
Let the area of the region { ( ?? , ?? ) : 0 = ?? = 3 , 0 = ?? = m i n { ?? 2
+ 2 , 2 ?? + 2 } } be ?? . Then 12 A
is equal to
Q8 - 2024 (30 Jan Shift 1)
The area (in square units) of the region bounded by the parabola y
2
= 4 ( x - 2 ) and the
line y = 2x - 8
(1) 8
(2) 9
(3) 6
(4) 7
Q9 - 2024 (30 Jan Shift 2)
Let ?? = ?? ( ?? ) be a curve lying in the first quadrant such that the area enclosed by the line
Y - y = Y
'
( x ) ( X - x ) and the co-ordinate axes, where ( ?? , ?? ) is any point on the curve, is
always
- ?? 2
2 Y
'
( x )
+ 1 , Y
'
( x ) ? 0. If Y ( 1 ) = 1, then 12Y ( 2 ) equals
Q10 - 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola ( ?? - 2 )
2
= ?? - 1, the line ?? - 2 ?? + 4 =
0 and the positive coordinate axes is
Q11 - 2024 (31 Jan Shift 1)
The area of the region
{ ( ?? , ?? ) : ?? 2
= 4 ?? , ?? < 4 ,
???? ( ?? - 1 ) ( ?? - 2 )
( ?? - 3 ) ( ?? - 4 )
> 0 , ?? ? 3 } is
(1)
16
3
(2)
64
3
(3)
8
3
(4)
32
3
Q12 - 2024 (31 Jan Shift 2)
The area of the region enclosed by the parabola ?? = 4 ?? - ?? 2
and 3 ?? = ( ?? - 4 )
2
is equal
to
(1)
32
9
(2) 4
(3) 6
(4)
14
3
Answer Key
Q1 (3) Q2 ( 7 ) Q3 (8) Q4 (119)
Q5 (304) Q6 (171) Q7 (164) Q8 (2)
Q9 (20) Q10 (5) Q11 (4) Q12 (3)
Solutions
Q1
???? + 4 ?? = 16 ? , ? ?? + ?? = 6
?? ( ?? + 4 ) = 16
x + y = 6
on solving, (1) \ & (2)
we get ?? = 4 , ?? = - 2
Page 4
JEE Mains Previous Year Questions
(2021-2024): Applications of Integrals
2024
Q1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves ???? + 4 ?? = 16 and ?? + ?? = 6 is equal to :
(1) 28 - 30 log
?? ? 2
(2) 30 - 28 log
?? ? 2
(3) 30 - 32 log
?? ? 2
(4) 32 - 30 log
?? ? 2
Q2 - 2024 (01 Feb Shift 2)
Three points O ( 0 , 0 ) , P ( a , a
2
) , Q ( - b , b
2
) , a > 0 , b > 0, are on the parabola ?? = ?? 2
. Let ?? 1
be the area of the region bounded by the line PQ and the parabola, and ?? 2
be the area
of the triangle ???? ?? . If the minimum value of
?? 1
?? 2
is
?? ?? , g c d ? ( ?? , ?? ) = 1, then ?? + ?? is equal
to:
Q3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of ?? , for which area of the region bounded by
the parabolas 2 y
2
= kx and ky
2
= 2 ( y - x ) is maximum, is equal to :
Q4 - 2024 (27 Jan Shift 1)
Let the area of the region { ( ?? , ?? ) : ?? - 2 ?? + 4 = 0 , ?? + 2 ?? 2
= 0 , ?? + 4 ?? 2
= 8 , ?? = 0 } be
?? ?? ,
where ?? and n are coprime numbers. Then m + n is equal to
Q5 - 2024 (27 Jan Shift 2)
If the area of the region { ( ?? , ?? ) : 0 = ?? = m i n { 2 ?? , 6 ?? - ?? 2
} } is ?? , then 12 A is equal to.
Q6 - 2024 (29 Jan Shift 1)
The area (in sq. units) of the part of circle ?? 2
+ ?? 2
= 169 which is below the line 5 ?? - ?? =
13 is
????
2 ?? -
65
2
+
?? ?? sin
- 1
? (
12
13
) where ?? , ?? are coprime numbers. Then ?? + ?? is equal to
Q7 - 2024 (29 Jan Shift 2)
Let the area of the region { ( ?? , ?? ) : 0 = ?? = 3 , 0 = ?? = m i n { ?? 2
+ 2 , 2 ?? + 2 } } be ?? . Then 12 A
is equal to
Q8 - 2024 (30 Jan Shift 1)
The area (in square units) of the region bounded by the parabola y
2
= 4 ( x - 2 ) and the
line y = 2x - 8
(1) 8
(2) 9
(3) 6
(4) 7
Q9 - 2024 (30 Jan Shift 2)
Let ?? = ?? ( ?? ) be a curve lying in the first quadrant such that the area enclosed by the line
Y - y = Y
'
( x ) ( X - x ) and the co-ordinate axes, where ( ?? , ?? ) is any point on the curve, is
always
- ?? 2
2 Y
'
( x )
+ 1 , Y
'
( x ) ? 0. If Y ( 1 ) = 1, then 12Y ( 2 ) equals
Q10 - 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola ( ?? - 2 )
2
= ?? - 1, the line ?? - 2 ?? + 4 =
0 and the positive coordinate axes is
Q11 - 2024 (31 Jan Shift 1)
The area of the region
{ ( ?? , ?? ) : ?? 2
= 4 ?? , ?? < 4 ,
???? ( ?? - 1 ) ( ?? - 2 )
( ?? - 3 ) ( ?? - 4 )
> 0 , ?? ? 3 } is
(1)
16
3
(2)
64
3
(3)
8
3
(4)
32
3
Q12 - 2024 (31 Jan Shift 2)
The area of the region enclosed by the parabola ?? = 4 ?? - ?? 2
and 3 ?? = ( ?? - 4 )
2
is equal
to
(1)
32
9
(2) 4
(3) 6
(4)
14
3
Answer Key
Q1 (3) Q2 ( 7 ) Q3 (8) Q4 (119)
Q5 (304) Q6 (171) Q7 (164) Q8 (2)
Q9 (20) Q10 (5) Q11 (4) Q12 (3)
Solutions
Q1
???? + 4 ?? = 16 ? , ? ?? + ?? = 6
?? ( ?? + 4 ) = 16
x + y = 6
on solving, (1) \ & (2)
we get ?? = 4 , ?? = - 2
Area = ? ?
4
- 2
? ( ( 6 - ?? ) - (
16
?? + 4
) ) ????
= 30 - 32ln ? 2
Q2
???? : - ?? - ?? 2
=
?? 2
- ?? 2
?? + ?? ( ?? - ?? )
?? - ?? 2
= ( ?? - ?? ) ?? - ( ?? - ?? ) ??
?? = ( ?? - ?? ) ?? + ????
?? 1
= ?
- ?? ?? ? ( ( ?? - ?? ) ?? + ???? - ?? 2
) ????
= ( ?? - ?? )
?? 2
2
+ ( ???? ) ?? -
?? 3
3
|
- ?? ??
Page 5
JEE Mains Previous Year Questions
(2021-2024): Applications of Integrals
2024
Q1 - 2024 (01 Feb Shift 1)
The area enclosed by the curves ???? + 4 ?? = 16 and ?? + ?? = 6 is equal to :
(1) 28 - 30 log
?? ? 2
(2) 30 - 28 log
?? ? 2
(3) 30 - 32 log
?? ? 2
(4) 32 - 30 log
?? ? 2
Q2 - 2024 (01 Feb Shift 2)
Three points O ( 0 , 0 ) , P ( a , a
2
) , Q ( - b , b
2
) , a > 0 , b > 0, are on the parabola ?? = ?? 2
. Let ?? 1
be the area of the region bounded by the line PQ and the parabola, and ?? 2
be the area
of the triangle ???? ?? . If the minimum value of
?? 1
?? 2
is
?? ?? , g c d ? ( ?? , ?? ) = 1, then ?? + ?? is equal
to:
Q3 - 2024 (01 Feb Shift 2)
The sum of squares of all possible values of ?? , for which area of the region bounded by
the parabolas 2 y
2
= kx and ky
2
= 2 ( y - x ) is maximum, is equal to :
Q4 - 2024 (27 Jan Shift 1)
Let the area of the region { ( ?? , ?? ) : ?? - 2 ?? + 4 = 0 , ?? + 2 ?? 2
= 0 , ?? + 4 ?? 2
= 8 , ?? = 0 } be
?? ?? ,
where ?? and n are coprime numbers. Then m + n is equal to
Q5 - 2024 (27 Jan Shift 2)
If the area of the region { ( ?? , ?? ) : 0 = ?? = m i n { 2 ?? , 6 ?? - ?? 2
} } is ?? , then 12 A is equal to.
Q6 - 2024 (29 Jan Shift 1)
The area (in sq. units) of the part of circle ?? 2
+ ?? 2
= 169 which is below the line 5 ?? - ?? =
13 is
????
2 ?? -
65
2
+
?? ?? sin
- 1
? (
12
13
) where ?? , ?? are coprime numbers. Then ?? + ?? is equal to
Q7 - 2024 (29 Jan Shift 2)
Let the area of the region { ( ?? , ?? ) : 0 = ?? = 3 , 0 = ?? = m i n { ?? 2
+ 2 , 2 ?? + 2 } } be ?? . Then 12 A
is equal to
Q8 - 2024 (30 Jan Shift 1)
The area (in square units) of the region bounded by the parabola y
2
= 4 ( x - 2 ) and the
line y = 2x - 8
(1) 8
(2) 9
(3) 6
(4) 7
Q9 - 2024 (30 Jan Shift 2)
Let ?? = ?? ( ?? ) be a curve lying in the first quadrant such that the area enclosed by the line
Y - y = Y
'
( x ) ( X - x ) and the co-ordinate axes, where ( ?? , ?? ) is any point on the curve, is
always
- ?? 2
2 Y
'
( x )
+ 1 , Y
'
( x ) ? 0. If Y ( 1 ) = 1, then 12Y ( 2 ) equals
Q10 - 2024 (30 Jan Shift 2)
The area of the region enclosed by the parabola ( ?? - 2 )
2
= ?? - 1, the line ?? - 2 ?? + 4 =
0 and the positive coordinate axes is
Q11 - 2024 (31 Jan Shift 1)
The area of the region
{ ( ?? , ?? ) : ?? 2
= 4 ?? , ?? < 4 ,
???? ( ?? - 1 ) ( ?? - 2 )
( ?? - 3 ) ( ?? - 4 )
> 0 , ?? ? 3 } is
(1)
16
3
(2)
64
3
(3)
8
3
(4)
32
3
Q12 - 2024 (31 Jan Shift 2)
The area of the region enclosed by the parabola ?? = 4 ?? - ?? 2
and 3 ?? = ( ?? - 4 )
2
is equal
to
(1)
32
9
(2) 4
(3) 6
(4)
14
3
Answer Key
Q1 (3) Q2 ( 7 ) Q3 (8) Q4 (119)
Q5 (304) Q6 (171) Q7 (164) Q8 (2)
Q9 (20) Q10 (5) Q11 (4) Q12 (3)
Solutions
Q1
???? + 4 ?? = 16 ? , ? ?? + ?? = 6
?? ( ?? + 4 ) = 16
x + y = 6
on solving, (1) \ & (2)
we get ?? = 4 , ?? = - 2
Area = ? ?
4
- 2
? ( ( 6 - ?? ) - (
16
?? + 4
) ) ????
= 30 - 32ln ? 2
Q2
???? : - ?? - ?? 2
=
?? 2
- ?? 2
?? + ?? ( ?? - ?? )
?? - ?? 2
= ( ?? - ?? ) ?? - ( ?? - ?? ) ??
?? = ( ?? - ?? ) ?? + ????
?? 1
= ?
- ?? ?? ? ( ( ?? - ?? ) ?? + ???? - ?? 2
) ????
= ( ?? - ?? )
?? 2
2
+ ( ???? ) ?? -
?? 3
3
|
- ?? ??
=
( ?? - ?? )
2
( ?? + ?? )
2
+ ???? ( ?? + ?? ) -
( ?? 3
+ ?? 3
)
3
?? 1
?? 2
=
( ?? - ?? )
2
2
+ ???? -
( ?? 2
+ ?? 2
- ???? )
3
athon
????
2
=
3 ( ?? - ?? )
2
+ 6 ???? - 2 ( ?? 2
+ ?? 2
- ???? )
3 ????
=
1
3
[
a
b
+
b
a
+ 2 ]
=
4
3
=
m
n
? m + n = 7
Q3
?? ?? 2
= 2 ( ?? - ?? )
2 y
2
= kx
Point of intersection ?
ky
2
= (
y - 2 y
2
k
)
y = 0 ? ky = 2 (
1 - 2y
k
)
ky +
4y
k
= 2
y =
2
k +
4
k
=
2k
k
2
+ 4
?? = ?
0
2 ?? ?? 2
+ 4
? ( ( ?? -
?? ?? 2
2
) - (
2 ?? 2
?? ) ) · ????
=
?? 2
2
- (
?? 2
+
2
?? ) ·
?? 3
3
|
0
2 ?? ?? 2
+ 4
= (
2 ?? ?? 2
+ 4
)
2
[
1
2
-
?? 2
+ 4
2 ?? ×
1
3
×
2 ?? ?? 2
+ 4
]
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FAQs on Application of Integrals: JEE Mains Previous Year Questions (2021-2024) - Mathematics (Maths) for JEE Main & Advanced
| 1. What is the application of integrals in JEE mains? |  |
Ans. The application of integrals in JEE mains involves using integration techniques to solve problems related to areas, volumes, and other physical quantities. It helps in finding the area under curves, calculating the displacement, and determining the rate of change of quantities.
| 2. How are integrals used to find the area under curves in JEE mains? | |
Ans. Integrals are used to find the area under curves in JEE mains by calculating the definite integral of a function over a given interval. The area under a curve can be determined by integrating the function and evaluating it within the specified interval.
| 3. Can integrals be used to calculate the volume of solids in JEE mains? | |
Ans. Yes, integrals can be used to calculate the volume of solids in JEE mains. By considering the cross-sections of a solid and integrating them over a given interval, the volume of the solid can be determined. This technique is known as the method of cross-sections.
| 4. How are integrals applied to calculate displacement in JEE mains? | |
Ans. Integrals are applied to calculate displacement in JEE mains by integrating the velocity function over a given interval. The result of the integration gives the change in position or displacement of an object over that interval.
| 5. How can integrals help in determining the rate of change of quantities in JEE mains? | |
Ans. Integrals can help in determining the rate of change of quantities in JEE mains by integrating the rate of change function over a given interval. The result of the integration gives the total change in the quantity over that interval, providing information about the rate of change.
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Application of Integrals: JEE Mains Previous Year Questions (2021-2024) JEE
The "Application of Integrals: JEE Mains Previous Year Questions (2021-2024) JEE Questions" guide is a valuable resource for all aspiring students preparing for the
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The content of Application of Integrals: JEE Mains Previous Year Questions (2021-2024) is prepared as per the latest JEE syllabus.
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