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Application of Integrals: JEE Main Previous Year Questions (2021-2026)
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Page 1
JEE Main Previous Year Questions
(2025): Application of Integrals
Question1: The area of the region, inside the circle ( ?? - ?? v ?? )
?? + ?? ?? = ???? and outside
the parabola ?? ?? = ?? v ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 3 ?? - 8
B. 6 ?? - 8
C. 3 ?? + 8
D. 6 ?? - 16
Answer: D
Solution:
Required area = 2 ?
0
2 v 3
? (
v
4 v 3 ?? - ?? 2
-
v
2 v 3 ?? ) ????
= 2 ?
0
2 v 3
? (
v
12 - ( ?? - 2 v 3 )
2
-
v
2 v 3 ?? ) ????
= 2 [
?? - 2 v 3
2
v
12 - ( ?? - 2 v 3 )
2
+
12
2
sin
- 1
? (
?? - 2 v 3
2 v 3
) -
v
2 v 3 ?? 3
2
3
2
]
0
2 v 3
= 2 { 3 ?? - 8 }
= 6 ?? - 16 sq. units.
Page 2
JEE Main Previous Year Questions
(2025): Application of Integrals
Question1: The area of the region, inside the circle ( ?? - ?? v ?? )
?? + ?? ?? = ???? and outside
the parabola ?? ?? = ?? v ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 3 ?? - 8
B. 6 ?? - 8
C. 3 ?? + 8
D. 6 ?? - 16
Answer: D
Solution:
Required area = 2 ?
0
2 v 3
? (
v
4 v 3 ?? - ?? 2
-
v
2 v 3 ?? ) ????
= 2 ?
0
2 v 3
? (
v
12 - ( ?? - 2 v 3 )
2
-
v
2 v 3 ?? ) ????
= 2 [
?? - 2 v 3
2
v
12 - ( ?? - 2 v 3 )
2
+
12
2
sin
- 1
? (
?? - 2 v 3
2 v 3
) -
v
2 v 3 ?? 3
2
3
2
]
0
2 v 3
= 2 { 3 ?? - 8 }
= 6 ?? - 16 sq. units.
Question2: The area of the region enclosed by the curves ?? = ?? ?? - ?? ?? + ?? and ?? ?? =
???? - ?? ?? is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
8
3
B. 5
C. 8
D.
4
3
Answer: A
Solution:
Consider the curves
?? = ?? 2
- 4 ?? + 4 = ( ?? - 2 )
2
and
?? 2
= 16 - 8 ?? .
Notice that the second equation can be rewritten in terms of ?? :
8 ?? = 16 - ?? 2
? ? ? ?? = 2 -
?? 2
8
.
Step 1. Find the Intersection Points
To find the points where the curves intersect, substitute
?? = ( ?? - 2 )
2
into
?? 2
= 16 - 8 ?? .
Let
?? = ?? - 2 ? so that ? ?? = ?? 2
.
Then
?? 2
= ?? 4
and
?? = ?? + 2.
Substitute into the second curve:
?? 4
= 16 - 8 ( ?? + 2 ) .
Simplify the right side:
16 - 8 ( ?? + 2 ) = 16 - 8 ?? - 16 = - 8 ?? .
Thus, the equation becomes
?? 4
+ 8 ?? = 0 ? ? ? ?? ( ?? 4
/ ?? ? * * correctingfactor * * ) .
In fact, factor by taking out a common factor ?? :
?? ( ?? 3
+ 8 ) = 0.
Thus, either
?? = 0 ? or ? ?? 3
= - 8.
For ?? = 0 :
Page 3
JEE Main Previous Year Questions
(2025): Application of Integrals
Question1: The area of the region, inside the circle ( ?? - ?? v ?? )
?? + ?? ?? = ???? and outside
the parabola ?? ?? = ?? v ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 3 ?? - 8
B. 6 ?? - 8
C. 3 ?? + 8
D. 6 ?? - 16
Answer: D
Solution:
Required area = 2 ?
0
2 v 3
? (
v
4 v 3 ?? - ?? 2
-
v
2 v 3 ?? ) ????
= 2 ?
0
2 v 3
? (
v
12 - ( ?? - 2 v 3 )
2
-
v
2 v 3 ?? ) ????
= 2 [
?? - 2 v 3
2
v
12 - ( ?? - 2 v 3 )
2
+
12
2
sin
- 1
? (
?? - 2 v 3
2 v 3
) -
v
2 v 3 ?? 3
2
3
2
]
0
2 v 3
= 2 { 3 ?? - 8 }
= 6 ?? - 16 sq. units.
Question2: The area of the region enclosed by the curves ?? = ?? ?? - ?? ?? + ?? and ?? ?? =
???? - ?? ?? is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
8
3
B. 5
C. 8
D.
4
3
Answer: A
Solution:
Consider the curves
?? = ?? 2
- 4 ?? + 4 = ( ?? - 2 )
2
and
?? 2
= 16 - 8 ?? .
Notice that the second equation can be rewritten in terms of ?? :
8 ?? = 16 - ?? 2
? ? ? ?? = 2 -
?? 2
8
.
Step 1. Find the Intersection Points
To find the points where the curves intersect, substitute
?? = ( ?? - 2 )
2
into
?? 2
= 16 - 8 ?? .
Let
?? = ?? - 2 ? so that ? ?? = ?? 2
.
Then
?? 2
= ?? 4
and
?? = ?? + 2.
Substitute into the second curve:
?? 4
= 16 - 8 ( ?? + 2 ) .
Simplify the right side:
16 - 8 ( ?? + 2 ) = 16 - 8 ?? - 16 = - 8 ?? .
Thus, the equation becomes
?? 4
+ 8 ?? = 0 ? ? ? ?? ( ?? 4
/ ?? ? * * correctingfactor * * ) .
In fact, factor by taking out a common factor ?? :
?? ( ?? 3
+ 8 ) = 0.
Thus, either
?? = 0 ? or ? ?? 3
= - 8.
For ?? = 0 :
?? = ?? + 2 = 2 , ?? = ?? 2
= 0.
For ?? 3
= - 8 :
?? = - 2 , ? so ? ?? = - 2 + 2 = 0 , ?? = ( - 2 )
2
= 4.
The curves intersect at the points ( 2 , 0 ) and ( 0 , 4 ) .
Step 2. Express the Curves in Terms of ??
It is easier to integrate horizontally by expressing ?? as a function of ?? .
From the second curve:
?? = 2 -
?? 2
8
.
From the first curve, solving
?? = ( ?? - 2 )
2
for ?? gives
?? - 2 = ± v ?? .
Since at the intersection ( 0 , 4 ) the ?? -value is less than 2 , we take the negative branch:
?? = 2 - v ?? .
Thus, for a fixed ?? between 0 and 4 , the left boundary is
?? left
= 2 - v ?? ,
and the right boundary is
?? right
= 2 -
?? 2
8
.
Step 3. Set Up the Integral for the Area
The horizontal distance between the curves at a given ?? is
? ?? = ?? right
- ?? left
= ( 2 -
?? 2
8
) - ( 2 - v ?? ) = v ?? -
?? 2
8
.
Integrate with respect to ?? from ?? = 0 to ?? = 4 :
?? = ?
0
4
? ( v ?? -
?? 2
8
) ???? .
Step 4. Evaluate the Integral
Write the integral as
?? = ?
0
4
? ?? 1 / 2
???? -
1
8
?
0
4
? ?? 2
???? .
Compute each term:
For the first integral:
? ?? 1 / 2
???? =
2
3
?? 3 / 2
.
For the second integral:
? ?? 2
???? =
?? 3
3
.
Thus,
?? = [
2
3
?? 3 / 2
]
0
4
-
1
8
[
?? 3
3
]
0
4
.
Substitute ?? = 4 (note that at ?? = 0 both terms vanish):
Compute 4
3 / 2
:
4
3 / 2
= ( v 4 )
3
= 2
3
= 8.
Compute the first term:
Page 4
JEE Main Previous Year Questions
(2025): Application of Integrals
Question1: The area of the region, inside the circle ( ?? - ?? v ?? )
?? + ?? ?? = ???? and outside
the parabola ?? ?? = ?? v ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 3 ?? - 8
B. 6 ?? - 8
C. 3 ?? + 8
D. 6 ?? - 16
Answer: D
Solution:
Required area = 2 ?
0
2 v 3
? (
v
4 v 3 ?? - ?? 2
-
v
2 v 3 ?? ) ????
= 2 ?
0
2 v 3
? (
v
12 - ( ?? - 2 v 3 )
2
-
v
2 v 3 ?? ) ????
= 2 [
?? - 2 v 3
2
v
12 - ( ?? - 2 v 3 )
2
+
12
2
sin
- 1
? (
?? - 2 v 3
2 v 3
) -
v
2 v 3 ?? 3
2
3
2
]
0
2 v 3
= 2 { 3 ?? - 8 }
= 6 ?? - 16 sq. units.
Question2: The area of the region enclosed by the curves ?? = ?? ?? - ?? ?? + ?? and ?? ?? =
???? - ?? ?? is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
8
3
B. 5
C. 8
D.
4
3
Answer: A
Solution:
Consider the curves
?? = ?? 2
- 4 ?? + 4 = ( ?? - 2 )
2
and
?? 2
= 16 - 8 ?? .
Notice that the second equation can be rewritten in terms of ?? :
8 ?? = 16 - ?? 2
? ? ? ?? = 2 -
?? 2
8
.
Step 1. Find the Intersection Points
To find the points where the curves intersect, substitute
?? = ( ?? - 2 )
2
into
?? 2
= 16 - 8 ?? .
Let
?? = ?? - 2 ? so that ? ?? = ?? 2
.
Then
?? 2
= ?? 4
and
?? = ?? + 2.
Substitute into the second curve:
?? 4
= 16 - 8 ( ?? + 2 ) .
Simplify the right side:
16 - 8 ( ?? + 2 ) = 16 - 8 ?? - 16 = - 8 ?? .
Thus, the equation becomes
?? 4
+ 8 ?? = 0 ? ? ? ?? ( ?? 4
/ ?? ? * * correctingfactor * * ) .
In fact, factor by taking out a common factor ?? :
?? ( ?? 3
+ 8 ) = 0.
Thus, either
?? = 0 ? or ? ?? 3
= - 8.
For ?? = 0 :
?? = ?? + 2 = 2 , ?? = ?? 2
= 0.
For ?? 3
= - 8 :
?? = - 2 , ? so ? ?? = - 2 + 2 = 0 , ?? = ( - 2 )
2
= 4.
The curves intersect at the points ( 2 , 0 ) and ( 0 , 4 ) .
Step 2. Express the Curves in Terms of ??
It is easier to integrate horizontally by expressing ?? as a function of ?? .
From the second curve:
?? = 2 -
?? 2
8
.
From the first curve, solving
?? = ( ?? - 2 )
2
for ?? gives
?? - 2 = ± v ?? .
Since at the intersection ( 0 , 4 ) the ?? -value is less than 2 , we take the negative branch:
?? = 2 - v ?? .
Thus, for a fixed ?? between 0 and 4 , the left boundary is
?? left
= 2 - v ?? ,
and the right boundary is
?? right
= 2 -
?? 2
8
.
Step 3. Set Up the Integral for the Area
The horizontal distance between the curves at a given ?? is
? ?? = ?? right
- ?? left
= ( 2 -
?? 2
8
) - ( 2 - v ?? ) = v ?? -
?? 2
8
.
Integrate with respect to ?? from ?? = 0 to ?? = 4 :
?? = ?
0
4
? ( v ?? -
?? 2
8
) ???? .
Step 4. Evaluate the Integral
Write the integral as
?? = ?
0
4
? ?? 1 / 2
???? -
1
8
?
0
4
? ?? 2
???? .
Compute each term:
For the first integral:
? ?? 1 / 2
???? =
2
3
?? 3 / 2
.
For the second integral:
? ?? 2
???? =
?? 3
3
.
Thus,
?? = [
2
3
?? 3 / 2
]
0
4
-
1
8
[
?? 3
3
]
0
4
.
Substitute ?? = 4 (note that at ?? = 0 both terms vanish):
Compute 4
3 / 2
:
4
3 / 2
= ( v 4 )
3
= 2
3
= 8.
Compute the first term:
2
3
× 8 =
16
3
.
Compute the second term:
1
8
·
64
3
=
64
24
=
8
3
.
Therefore, the area is
?? =
16
3
-
8
3
=
8
3
.
Final Answer
The area of the region enclosed by the curves is
8
3
.
Question3: If the area of the region
{ ( ?? , ?? ) : - ?? = ?? = ?? , ?? = ?? = ?? + ?? | ?? |
- ?? - ?? , ?? > ?? } is
?? ?? + ???? + ?? ?? , then the value of ?? is :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 7
B. 5
C. 6
D. 8
Answer: B
Solution:
Page 5
JEE Main Previous Year Questions
(2025): Application of Integrals
Question1: The area of the region, inside the circle ( ?? - ?? v ?? )
?? + ?? ?? = ???? and outside
the parabola ?? ?? = ?? v ?? ?? is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 3 ?? - 8
B. 6 ?? - 8
C. 3 ?? + 8
D. 6 ?? - 16
Answer: D
Solution:
Required area = 2 ?
0
2 v 3
? (
v
4 v 3 ?? - ?? 2
-
v
2 v 3 ?? ) ????
= 2 ?
0
2 v 3
? (
v
12 - ( ?? - 2 v 3 )
2
-
v
2 v 3 ?? ) ????
= 2 [
?? - 2 v 3
2
v
12 - ( ?? - 2 v 3 )
2
+
12
2
sin
- 1
? (
?? - 2 v 3
2 v 3
) -
v
2 v 3 ?? 3
2
3
2
]
0
2 v 3
= 2 { 3 ?? - 8 }
= 6 ?? - 16 sq. units.
Question2: The area of the region enclosed by the curves ?? = ?? ?? - ?? ?? + ?? and ?? ?? =
???? - ?? ?? is:
JEE Main 2025 (Online) 22nd January Evening Shift
Options:
A.
8
3
B. 5
C. 8
D.
4
3
Answer: A
Solution:
Consider the curves
?? = ?? 2
- 4 ?? + 4 = ( ?? - 2 )
2
and
?? 2
= 16 - 8 ?? .
Notice that the second equation can be rewritten in terms of ?? :
8 ?? = 16 - ?? 2
? ? ? ?? = 2 -
?? 2
8
.
Step 1. Find the Intersection Points
To find the points where the curves intersect, substitute
?? = ( ?? - 2 )
2
into
?? 2
= 16 - 8 ?? .
Let
?? = ?? - 2 ? so that ? ?? = ?? 2
.
Then
?? 2
= ?? 4
and
?? = ?? + 2.
Substitute into the second curve:
?? 4
= 16 - 8 ( ?? + 2 ) .
Simplify the right side:
16 - 8 ( ?? + 2 ) = 16 - 8 ?? - 16 = - 8 ?? .
Thus, the equation becomes
?? 4
+ 8 ?? = 0 ? ? ? ?? ( ?? 4
/ ?? ? * * correctingfactor * * ) .
In fact, factor by taking out a common factor ?? :
?? ( ?? 3
+ 8 ) = 0.
Thus, either
?? = 0 ? or ? ?? 3
= - 8.
For ?? = 0 :
?? = ?? + 2 = 2 , ?? = ?? 2
= 0.
For ?? 3
= - 8 :
?? = - 2 , ? so ? ?? = - 2 + 2 = 0 , ?? = ( - 2 )
2
= 4.
The curves intersect at the points ( 2 , 0 ) and ( 0 , 4 ) .
Step 2. Express the Curves in Terms of ??
It is easier to integrate horizontally by expressing ?? as a function of ?? .
From the second curve:
?? = 2 -
?? 2
8
.
From the first curve, solving
?? = ( ?? - 2 )
2
for ?? gives
?? - 2 = ± v ?? .
Since at the intersection ( 0 , 4 ) the ?? -value is less than 2 , we take the negative branch:
?? = 2 - v ?? .
Thus, for a fixed ?? between 0 and 4 , the left boundary is
?? left
= 2 - v ?? ,
and the right boundary is
?? right
= 2 -
?? 2
8
.
Step 3. Set Up the Integral for the Area
The horizontal distance between the curves at a given ?? is
? ?? = ?? right
- ?? left
= ( 2 -
?? 2
8
) - ( 2 - v ?? ) = v ?? -
?? 2
8
.
Integrate with respect to ?? from ?? = 0 to ?? = 4 :
?? = ?
0
4
? ( v ?? -
?? 2
8
) ???? .
Step 4. Evaluate the Integral
Write the integral as
?? = ?
0
4
? ?? 1 / 2
???? -
1
8
?
0
4
? ?? 2
???? .
Compute each term:
For the first integral:
? ?? 1 / 2
???? =
2
3
?? 3 / 2
.
For the second integral:
? ?? 2
???? =
?? 3
3
.
Thus,
?? = [
2
3
?? 3 / 2
]
0
4
-
1
8
[
?? 3
3
]
0
4
.
Substitute ?? = 4 (note that at ?? = 0 both terms vanish):
Compute 4
3 / 2
:
4
3 / 2
= ( v 4 )
3
= 2
3
= 8.
Compute the first term:
2
3
× 8 =
16
3
.
Compute the second term:
1
8
·
64
3
=
64
24
=
8
3
.
Therefore, the area is
?? =
16
3
-
8
3
=
8
3
.
Final Answer
The area of the region enclosed by the curves is
8
3
.
Question3: If the area of the region
{ ( ?? , ?? ) : - ?? = ?? = ?? , ?? = ?? = ?? + ?? | ?? |
- ?? - ?? , ?? > ?? } is
?? ?? + ???? + ?? ?? , then the value of ?? is :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 7
B. 5
C. 6
D. 8
Answer: B
Solution:
required area is ?? + ?
0
1
? ( ?? + ?? ?? - ?? - ?? ) ????
?? + [ ?? + ?? ?? + ?? - ?? ]
0
1
2 ?? + ?? - 1 + ?? - 1
- 1 = ?? + 8 +
1
??
2 ?? = 10 ? ?? = 5
Question4: The area of the region { ( ?? , ?? ) : ?? ?? + ?? ?? + ?? = ?? = | ?? + ?? | } is equal to
JEE Main 2025 (Online) 24th January Morning Shift
Options:
A. 7
B. 24 / 5
C. 20 / 3
D. 5
Answer: C
Solution:
?? 2
+ 4 ?? + 2 = ?? = | ?? + 2 |
The area bounded between
?? = ?? 2
+ 4 ?? + 2 = ( ?? + 2 )
2
- 2
and ?? = | ?? + 2 | is same as area bounded between ?? = ?? 2
- 2 and ?? = | ?? |
For P.O.I | ?? |
2
- 2 = | ?? |
? | ?? | = 2 ? ?? = ± 2
? Required area = - ?
- 2
2
? ( ?? 2
- 2 ) ???? + ?
- 2
2
? | ?? | ????
= - 2 ?
0
2
? ( ?? 2
- 2 ) ???? + 2 ?
0
2
? ?? · ????
= - 2 [
?? 3
3
- 2 ?? ]
0
2
+ 2 [
?? 2
2
]
0
2
= - - 2 [
8
3
- 4 ] + 2 [
4
2
]
= - 2 × (
- 4
3
) + 4
=
20
3
Question5: The area of the region enclosed by the curves ?? = ?? ?? , ?? = | ?? ?? - ?? | and ?? -
axis is :
JEE Main 2025 (Online) 24th January Evening Shift
Options:
Read MoreFAQs on Application of Integrals: JEE Main Previous Year Questions (2021-2026)
| 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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