Detailed Notes: Area Under the curve | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


AREA UNDER THE CURVE 
Understanding the area under curves is a fundamental concept in calculus that finds applications 
in various fields such as physics, engineering, and economics. This topic allows us to explore how 
integral calculus helps in determining the area bounded by curves, lines, and axes, offering a 
powerful tool for solving complex problems. 
1. AREA UNDER THE CURVES : 
Let’s see how to calculate the area under different types of curves. 
(a) Area bounded by the curve ?? = ?? ( ?? ) , the ?? -axis and the lines ?? = a and ?? = ?? is given by ?? =
?
?? ?? ??????? , where ?? = ?? ( ?? ) lies above the ?? -axis and ?? > ?? . Here vertical strip of thickness ???? is 
considered at distance ?? . 
 
(b) If ?? = ?? ( ?? ) lies completely below the ?? -axis then ?? is negative and we consider the magnitude 
only, i.e. ?? = |?
?? ?? ??????? | 
 
Page 2


AREA UNDER THE CURVE 
Understanding the area under curves is a fundamental concept in calculus that finds applications 
in various fields such as physics, engineering, and economics. This topic allows us to explore how 
integral calculus helps in determining the area bounded by curves, lines, and axes, offering a 
powerful tool for solving complex problems. 
1. AREA UNDER THE CURVES : 
Let’s see how to calculate the area under different types of curves. 
(a) Area bounded by the curve ?? = ?? ( ?? ) , the ?? -axis and the lines ?? = a and ?? = ?? is given by ?? =
?
?? ?? ??????? , where ?? = ?? ( ?? ) lies above the ?? -axis and ?? > ?? . Here vertical strip of thickness ???? is 
considered at distance ?? . 
 
(b) If ?? = ?? ( ?? ) lies completely below the ?? -axis then ?? is negative and we consider the magnitude 
only, i.e. ?? = |?
?? ?? ??????? | 
 
(c) If curve crosses the ?? -axis at ?? = ?? , then ?? = |?
?? ?? ??????? | + ?
?? ?? ??????? 
 
(d) Sometimes integration w.r.t. y is very useful (horizontal strip) : 
Area bounded by the curve, ?? -axis and the two abscissae at 
 
 
?? = ?? & ?? = ?? is written as ?? = ?
?? ?? ??????? . 
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = ?? 
(Area of one symmetric portion). 
Problem 1 : Find the area bounded by ?? = ?????? 2
 ?? , ?? =
?? 6
, ?? =
?? 3
&?? -axis 
Solution:   Area bounded = ?
?? /6
?? /3
??????? = ?
?? /6
?? /3
??????? 2
 ?????? = [?????? ?? ]
?? /6
?? /3
= ?????? 
?? 3
- ?????? 
?? 6
= v 3 -
1
v 3
=
2
v 3
 
sq.units. 
Problem 2: Find the area in the first quadrant bounded by ?? = 4?? 2
, ?? = 0, ?? = 1 and ?? = 4. 
Solution :   Required area  = ?
1
4
??????? = ?
1
4
?
v ?? 2
???? =
1
2
[
2
3
?? 3/2
]
1
4
 
=
1
3
[4
3/2
- 1] =
1
3
[8 - 1] 
Page 3


AREA UNDER THE CURVE 
Understanding the area under curves is a fundamental concept in calculus that finds applications 
in various fields such as physics, engineering, and economics. This topic allows us to explore how 
integral calculus helps in determining the area bounded by curves, lines, and axes, offering a 
powerful tool for solving complex problems. 
1. AREA UNDER THE CURVES : 
Let’s see how to calculate the area under different types of curves. 
(a) Area bounded by the curve ?? = ?? ( ?? ) , the ?? -axis and the lines ?? = a and ?? = ?? is given by ?? =
?
?? ?? ??????? , where ?? = ?? ( ?? ) lies above the ?? -axis and ?? > ?? . Here vertical strip of thickness ???? is 
considered at distance ?? . 
 
(b) If ?? = ?? ( ?? ) lies completely below the ?? -axis then ?? is negative and we consider the magnitude 
only, i.e. ?? = |?
?? ?? ??????? | 
 
(c) If curve crosses the ?? -axis at ?? = ?? , then ?? = |?
?? ?? ??????? | + ?
?? ?? ??????? 
 
(d) Sometimes integration w.r.t. y is very useful (horizontal strip) : 
Area bounded by the curve, ?? -axis and the two abscissae at 
 
 
?? = ?? & ?? = ?? is written as ?? = ?
?? ?? ??????? . 
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = ?? 
(Area of one symmetric portion). 
Problem 1 : Find the area bounded by ?? = ?????? 2
 ?? , ?? =
?? 6
, ?? =
?? 3
&?? -axis 
Solution:   Area bounded = ?
?? /6
?? /3
??????? = ?
?? /6
?? /3
??????? 2
 ?????? = [?????? ?? ]
?? /6
?? /3
= ?????? 
?? 3
- ?????? 
?? 6
= v 3 -
1
v 3
=
2
v 3
 
sq.units. 
Problem 2: Find the area in the first quadrant bounded by ?? = 4?? 2
, ?? = 0, ?? = 1 and ?? = 4. 
Solution :   Required area  = ?
1
4
??????? = ?
1
4
?
v ?? 2
???? =
1
2
[
2
3
?? 3/2
]
1
4
 
=
1
3
[4
3/2
- 1] =
1
3
[8 - 1] 
 
=
7
3
= 2
1
3
 sq.units. 
Problem 3 : Find the area bounded by the curve ?? = ?????? 2?? , ?? -axis and the lines ?? =
?? 4
 and ?? =
3?? 4
 
Solution :   Required area = ?
?? /4
?? /2
??????? 2?????? + |?
?? /2
3?? /4
??????? 2?????? | = (-
?????? 2?? 2
)|
?? /4
?? /2
+ |(-
?????? 2?? 2
)|
?? /2
3?? /4
 
= -
1
2
[-1 - 0] + |
1
2
( 0 + ( -1) ) | = 1 ???? . ????????  
 
2. AREA ENCLOSED BETWEEN TWO 
CURVES : 
In the previous section, we saw how to calculate area between curve and axis of the coordinate 
plane. In this section, we will see how to calculate area between two different curves. 
Page 4


AREA UNDER THE CURVE 
Understanding the area under curves is a fundamental concept in calculus that finds applications 
in various fields such as physics, engineering, and economics. This topic allows us to explore how 
integral calculus helps in determining the area bounded by curves, lines, and axes, offering a 
powerful tool for solving complex problems. 
1. AREA UNDER THE CURVES : 
Let’s see how to calculate the area under different types of curves. 
(a) Area bounded by the curve ?? = ?? ( ?? ) , the ?? -axis and the lines ?? = a and ?? = ?? is given by ?? =
?
?? ?? ??????? , where ?? = ?? ( ?? ) lies above the ?? -axis and ?? > ?? . Here vertical strip of thickness ???? is 
considered at distance ?? . 
 
(b) If ?? = ?? ( ?? ) lies completely below the ?? -axis then ?? is negative and we consider the magnitude 
only, i.e. ?? = |?
?? ?? ??????? | 
 
(c) If curve crosses the ?? -axis at ?? = ?? , then ?? = |?
?? ?? ??????? | + ?
?? ?? ??????? 
 
(d) Sometimes integration w.r.t. y is very useful (horizontal strip) : 
Area bounded by the curve, ?? -axis and the two abscissae at 
 
 
?? = ?? & ?? = ?? is written as ?? = ?
?? ?? ??????? . 
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = ?? 
(Area of one symmetric portion). 
Problem 1 : Find the area bounded by ?? = ?????? 2
 ?? , ?? =
?? 6
, ?? =
?? 3
&?? -axis 
Solution:   Area bounded = ?
?? /6
?? /3
??????? = ?
?? /6
?? /3
??????? 2
 ?????? = [?????? ?? ]
?? /6
?? /3
= ?????? 
?? 3
- ?????? 
?? 6
= v 3 -
1
v 3
=
2
v 3
 
sq.units. 
Problem 2: Find the area in the first quadrant bounded by ?? = 4?? 2
, ?? = 0, ?? = 1 and ?? = 4. 
Solution :   Required area  = ?
1
4
??????? = ?
1
4
?
v ?? 2
???? =
1
2
[
2
3
?? 3/2
]
1
4
 
=
1
3
[4
3/2
- 1] =
1
3
[8 - 1] 
 
=
7
3
= 2
1
3
 sq.units. 
Problem 3 : Find the area bounded by the curve ?? = ?????? 2?? , ?? -axis and the lines ?? =
?? 4
 and ?? =
3?? 4
 
Solution :   Required area = ?
?? /4
?? /2
??????? 2?????? + |?
?? /2
3?? /4
??????? 2?????? | = (-
?????? 2?? 2
)|
?? /4
?? /2
+ |(-
?????? 2?? 2
)|
?? /2
3?? /4
 
= -
1
2
[-1 - 0] + |
1
2
( 0 + ( -1) ) | = 1 ???? . ????????  
 
2. AREA ENCLOSED BETWEEN TWO 
CURVES : 
In the previous section, we saw how to calculate area between curve and axis of the coordinate 
plane. In this section, we will see how to calculate area between two different curves. 
(a) Area bounded by two curves ?? = ?? (x) & ?? = ?? ( ?? ) such that ?? ( ?? )> ?? ( ?? ) is 
 
?? = ?
?? 1
?? 2
?( ?? 1
- ?? 2
) ???? 
?? = ?
?? 1
?? 2
?[?? ( ?? )- ?? ( ?? ) ]???? 
(b) In case horizontal strip is taken we have 
 ?? = ? ?
?? 2
?? 1
??( ?? 1
- ?? 2
) ????  ?? = ? ?
?? 2
?? 1
??[?? ( ?? )- ?? ( ?? ) ]????  
 
(c) If the curves ?? 1
= ?? ( ?? ) and ?? 2
= ?? ( ?? ) intersect at ?? = ?? , then required area 
?? = ? ?
?? ?? ?( ?? ( ?? )- ?? ( ?? ) ) ???? + ? ?
?? ?? ?( ?? ( ?? )- ?? ( ?? ) ) ???? = ? ?
?? ?? ?|?? ( ?? )- ?? ( ?? ) |???? 
Page 5


AREA UNDER THE CURVE 
Understanding the area under curves is a fundamental concept in calculus that finds applications 
in various fields such as physics, engineering, and economics. This topic allows us to explore how 
integral calculus helps in determining the area bounded by curves, lines, and axes, offering a 
powerful tool for solving complex problems. 
1. AREA UNDER THE CURVES : 
Let’s see how to calculate the area under different types of curves. 
(a) Area bounded by the curve ?? = ?? ( ?? ) , the ?? -axis and the lines ?? = a and ?? = ?? is given by ?? =
?
?? ?? ??????? , where ?? = ?? ( ?? ) lies above the ?? -axis and ?? > ?? . Here vertical strip of thickness ???? is 
considered at distance ?? . 
 
(b) If ?? = ?? ( ?? ) lies completely below the ?? -axis then ?? is negative and we consider the magnitude 
only, i.e. ?? = |?
?? ?? ??????? | 
 
(c) If curve crosses the ?? -axis at ?? = ?? , then ?? = |?
?? ?? ??????? | + ?
?? ?? ??????? 
 
(d) Sometimes integration w.r.t. y is very useful (horizontal strip) : 
Area bounded by the curve, ?? -axis and the two abscissae at 
 
 
?? = ?? & ?? = ?? is written as ?? = ?
?? ?? ??????? . 
Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = ?? 
(Area of one symmetric portion). 
Problem 1 : Find the area bounded by ?? = ?????? 2
 ?? , ?? =
?? 6
, ?? =
?? 3
&?? -axis 
Solution:   Area bounded = ?
?? /6
?? /3
??????? = ?
?? /6
?? /3
??????? 2
 ?????? = [?????? ?? ]
?? /6
?? /3
= ?????? 
?? 3
- ?????? 
?? 6
= v 3 -
1
v 3
=
2
v 3
 
sq.units. 
Problem 2: Find the area in the first quadrant bounded by ?? = 4?? 2
, ?? = 0, ?? = 1 and ?? = 4. 
Solution :   Required area  = ?
1
4
??????? = ?
1
4
?
v ?? 2
???? =
1
2
[
2
3
?? 3/2
]
1
4
 
=
1
3
[4
3/2
- 1] =
1
3
[8 - 1] 
 
=
7
3
= 2
1
3
 sq.units. 
Problem 3 : Find the area bounded by the curve ?? = ?????? 2?? , ?? -axis and the lines ?? =
?? 4
 and ?? =
3?? 4
 
Solution :   Required area = ?
?? /4
?? /2
??????? 2?????? + |?
?? /2
3?? /4
??????? 2?????? | = (-
?????? 2?? 2
)|
?? /4
?? /2
+ |(-
?????? 2?? 2
)|
?? /2
3?? /4
 
= -
1
2
[-1 - 0] + |
1
2
( 0 + ( -1) ) | = 1 ???? . ????????  
 
2. AREA ENCLOSED BETWEEN TWO 
CURVES : 
In the previous section, we saw how to calculate area between curve and axis of the coordinate 
plane. In this section, we will see how to calculate area between two different curves. 
(a) Area bounded by two curves ?? = ?? (x) & ?? = ?? ( ?? ) such that ?? ( ?? )> ?? ( ?? ) is 
 
?? = ?
?? 1
?? 2
?( ?? 1
- ?? 2
) ???? 
?? = ?
?? 1
?? 2
?[?? ( ?? )- ?? ( ?? ) ]???? 
(b) In case horizontal strip is taken we have 
 ?? = ? ?
?? 2
?? 1
??( ?? 1
- ?? 2
) ????  ?? = ? ?
?? 2
?? 1
??[?? ( ?? )- ?? ( ?? ) ]????  
 
(c) If the curves ?? 1
= ?? ( ?? ) and ?? 2
= ?? ( ?? ) intersect at ?? = ?? , then required area 
?? = ? ?
?? ?? ?( ?? ( ?? )- ?? ( ?? ) ) ???? + ? ?
?? ?? ?( ?? ( ?? )- ?? ( ?? ) ) ???? = ? ?
?? ?? ?|?? ( ?? )- ?? ( ?? ) |???? 
 
Note : Required area must have all the boundaries indicated in the problem. 
Problem 4: Find the area bounded by the curve ?? = ( ?? - 1) ( ?? - 2) ( ?? - 3) lying between the 
ordinates ?? = 0 and ?? = 3 and ?? -axis 
Solution : To determine the sign, we follow the usual rule of change of sign. 
?? = +????  ?????? ?? > 3 ?? = - ????   ?????? 2 < ?? < 3 ?? = + ????   ?????? 1 < ?? < 2 ?? = - ????   ?????? ?? < 1 ? ?
3
0
??|?? |???? = ? ?
1
0
??|?? |???? + ? ?
2
1
??|?? |???? + ? ?
3
2
??|?? |????  
= ? ?
1
0
??- ?????? + ? ?
2
1
???????? + ? ?
3
2
??- ??????  
 
Now let ?? ( ?? ) = ?( ?? - 1) ( ?? - 2) ( ?? - 3) ???? = ?( ?? 3
- 6?? 2
+ 11?? - 6) ???? =
1
4
?? 4
- 2?? 3
+
11
2
?? 2
- 6?? . 
? ?? ( 0) = 0, ?? ( 1)= -
9
4
, ?? ( 2) = -2, ?? ( 3) = -
9
4
. 
Hence required Area = -[?? ( 1)- ?? ( 0) ] + [?? ( 2)- ?? ( 1) ] - [?? ( 3)- ?? ( 2) ] = 2
3
4
 sq.units. 
Problem 5 : Compute the area of the figure bounded by the straight lines ?? = 0, ?? = 2 and the 
curves ?? = 2
?? , ?? = 2?? - ?? 2
.