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JEE Main Previous Year Questions (2026): Application of Integrals
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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 JEE Main Previous Year Questions (2026): Application of Integrals
| 1. What is the significance of the application of integrals in solving real-world problems? |  |
Ans. The application of integrals is crucial in various fields such as physics, engineering, and economics. Integrals help in calculating areas under curves, volumes of solids of revolution, and work done by forces. For example, in physics, integrals are used to determine the center of mass and the electric field generated by continuous charge distributions. This practical utility highlights the importance of mastering integrals in mathematical studies.
| 2. How can one find the area between two curves using integrals? | |
Ans. To find the area between two curves, one must first identify the points of intersection of the curves. Once the limits of integration are established, the area can be calculated by integrating the difference of the two functions. Mathematically, if you have two functions f(x) and g(x) where f(x) ≥ g(x) between x = a and x = b, the area A can be expressed as A = ∫[a to b] (f(x) - g(x)) dx. This method allows for accurate area determination between the curves.
| 3. What are the common techniques for solving integral problems in exams? | |
Ans. Common techniques for solving integral problems include substitution, integration by parts, and partial fraction decomposition. Substitution is used to simplify integrals by changing variables. Integration by parts is useful for integrating products of functions. Partial fraction decomposition helps in breaking down rational functions into simpler fractions that are easier to integrate. Mastery of these techniques is essential for efficient problem-solving in exams.
| 4. Can you explain the concept of definite and indefinite integrals? | |
Ans. Definite integrals have specific upper and lower limits and yield a numerical value representing the area under the curve between those limits. In contrast, indefinite integrals do not have limits and represent a family of functions plus a constant of integration (C). For instance, the integral of f(x) dx = F(x) + C, where F(x) is the antiderivative of f(x). Understanding the distinction between these two types of integrals is fundamental in calculus.
| 5. What role does the Fundamental Theorem of Calculus play in the application of integrals? | |
Ans. The Fundamental Theorem of Calculus connects differentiation and integration, providing a powerful tool for evaluating definite integrals. It states that if F is an antiderivative of a continuous function f on an interval [a, b], then the definite integral of f from a to b is given by F(b) - F(a). This theorem simplifies the process of finding areas and accumulated quantities, making it a cornerstone of integral calculus.
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