Detailed Notes: Definite Integral | Mathematics (Maths) for JEE Main & Advanced PDF Download

 Page 1


DEFINITE INTEGRATION 
Imagine looking at the shadow cast by a tree at sunset. The length and shape of the shadow vary with 
the terrain and the tree's shape. To find the exact area of that shadow, we use the concept of the definite 
integral —a mathematical tool that calculates the area under a curve. 
A definite integral, written as ?
?? ?? ??? ( ?? ) ???? , measures the total area between a curve ?? ( ?? ) and the x-axis 
from ?? = ?? , ?? = ?? . This concept is crucial in fields like finance, where it helps quantify continuous 
change, track investments, and analyze economic trends, making it invaluable for informed decision-
making. 
1. THE FUNDAMENTAL THEOREM OF 
CALCULUS: 
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection 
between the two branches of calculus: differential calculus and integral calculus. Differential calculus 
arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, 
the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two 
problems are actually closely related. In fact, he realized that differentiation and integration are inverse 
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the 
derivative and the integral. It was Newton and Leibnitz who exploited this relationship and used it to 
develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental 
Theorem enabled them to compute areas and integrals very easily without having to compute them as 
limits of sums. 
 
The Fundamental Theorem of Calculus, Part 1 
If ?? is continuous on [?? , ?? ], then the function ?? defined by 
?? ( ?? ) = ? ?
?? ?? ??? ( ?? ) ???? ?? = ?? = ?? 
Page 2


DEFINITE INTEGRATION 
Imagine looking at the shadow cast by a tree at sunset. The length and shape of the shadow vary with 
the terrain and the tree's shape. To find the exact area of that shadow, we use the concept of the definite 
integral —a mathematical tool that calculates the area under a curve. 
A definite integral, written as ?
?? ?? ??? ( ?? ) ???? , measures the total area between a curve ?? ( ?? ) and the x-axis 
from ?? = ?? , ?? = ?? . This concept is crucial in fields like finance, where it helps quantify continuous 
change, track investments, and analyze economic trends, making it invaluable for informed decision-
making. 
1. THE FUNDAMENTAL THEOREM OF 
CALCULUS: 
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection 
between the two branches of calculus: differential calculus and integral calculus. Differential calculus 
arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, 
the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two 
problems are actually closely related. In fact, he realized that differentiation and integration are inverse 
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the 
derivative and the integral. It was Newton and Leibnitz who exploited this relationship and used it to 
develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental 
Theorem enabled them to compute areas and integrals very easily without having to compute them as 
limits of sums. 
 
The Fundamental Theorem of Calculus, Part 1 
If ?? is continuous on [?? , ?? ], then the function ?? defined by 
?? ( ?? ) = ? ?
?? ?? ??? ( ?? ) ???? ?? = ?? = ?? 
is continuous on [?? , ?? ] and differentiable on ( ?? , ?? ) , and ?? '
( ?? ) = ?? ( ?? ) . 
The Fundamental Theorem of Calculus, Part 2 
If ?? is continuous on [?? , ?? ], then 
? ?
?? ?? ??? ( ?? ) ???? = ?? ( ?? )- ?? ( ?? ) 
where ?? is any antiderivative of ?? , that is, a function such that ?? '
= ?? . 
Note: If ?
?? ?? ??? ( ?? ) ???? = 0 ? then the equation ?? ( ?? ) = 0 has atleast one root lying in ( ?? , ?? ) provided ?? is a 
continuous function in ( ?? , ?? ) . 
2. PROPERTIES OF DEFINITE INTEGRAL: 
Some properties which are quite useful while solving any kind of questions are: 
(a) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? provided ?? is same 
(b) ?
?? ?? ??? ( ?? ) ???? = -?
?? ?? ??? ( ?? ) ???? 
(c) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? + ?
?? ?? ??? ( ?? ) ???? , where c may lie inside or outside the interval [?? , ?? ]. This 
property is to be used when ?? is piecewise continuous in ( ?? , ?? ) . 
Problem1: If ?? ( ?? ) = {?? 2
, 0 < ?? < 2 3?? - 4,2 = ?? < 3  then evaluate ?
0
3
??? ( ?? ) ???? 
Solution:  ?
0
3
??? ( ?? ) ???? = ?
0
2
??? ( ?? ) ???? + ?
2
3
??? ( ?? ) ???? = ?
0
2
??? 2
???? + ?
2
3
?( 3?? - 4) ???? 
= (
?? 3
3
)
0
2
+ (
3?? 2
2
- 4?? )
2
3
=
8
3
+
27
2
- 12 - 6 + 8 = 37/6#( ?????? . )  
 
 
function) 
(A) -
11
2
 
(B) -
7
2
 
(C) -6 
(D) -
17
2
 
Solution:  3[?? ] - 5
|?? |
?? = 3[?? ] - 5, if ?? > 0 
= 3[?? ] + 5, if ?? < 0 
Page 3


DEFINITE INTEGRATION 
Imagine looking at the shadow cast by a tree at sunset. The length and shape of the shadow vary with 
the terrain and the tree's shape. To find the exact area of that shadow, we use the concept of the definite 
integral —a mathematical tool that calculates the area under a curve. 
A definite integral, written as ?
?? ?? ??? ( ?? ) ???? , measures the total area between a curve ?? ( ?? ) and the x-axis 
from ?? = ?? , ?? = ?? . This concept is crucial in fields like finance, where it helps quantify continuous 
change, track investments, and analyze economic trends, making it invaluable for informed decision-
making. 
1. THE FUNDAMENTAL THEOREM OF 
CALCULUS: 
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection 
between the two branches of calculus: differential calculus and integral calculus. Differential calculus 
arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, 
the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two 
problems are actually closely related. In fact, he realized that differentiation and integration are inverse 
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the 
derivative and the integral. It was Newton and Leibnitz who exploited this relationship and used it to 
develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental 
Theorem enabled them to compute areas and integrals very easily without having to compute them as 
limits of sums. 
 
The Fundamental Theorem of Calculus, Part 1 
If ?? is continuous on [?? , ?? ], then the function ?? defined by 
?? ( ?? ) = ? ?
?? ?? ??? ( ?? ) ???? ?? = ?? = ?? 
is continuous on [?? , ?? ] and differentiable on ( ?? , ?? ) , and ?? '
( ?? ) = ?? ( ?? ) . 
The Fundamental Theorem of Calculus, Part 2 
If ?? is continuous on [?? , ?? ], then 
? ?
?? ?? ??? ( ?? ) ???? = ?? ( ?? )- ?? ( ?? ) 
where ?? is any antiderivative of ?? , that is, a function such that ?? '
= ?? . 
Note: If ?
?? ?? ??? ( ?? ) ???? = 0 ? then the equation ?? ( ?? ) = 0 has atleast one root lying in ( ?? , ?? ) provided ?? is a 
continuous function in ( ?? , ?? ) . 
2. PROPERTIES OF DEFINITE INTEGRAL: 
Some properties which are quite useful while solving any kind of questions are: 
(a) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? provided ?? is same 
(b) ?
?? ?? ??? ( ?? ) ???? = -?
?? ?? ??? ( ?? ) ???? 
(c) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? + ?
?? ?? ??? ( ?? ) ???? , where c may lie inside or outside the interval [?? , ?? ]. This 
property is to be used when ?? is piecewise continuous in ( ?? , ?? ) . 
Problem1: If ?? ( ?? ) = {?? 2
, 0 < ?? < 2 3?? - 4,2 = ?? < 3  then evaluate ?
0
3
??? ( ?? ) ???? 
Solution:  ?
0
3
??? ( ?? ) ???? = ?
0
2
??? ( ?? ) ???? + ?
2
3
??? ( ?? ) ???? = ?
0
2
??? 2
???? + ?
2
3
?( 3?? - 4) ???? 
= (
?? 3
3
)
0
2
+ (
3?? 2
2
- 4?? )
2
3
=
8
3
+
27
2
- 12 - 6 + 8 = 37/6#( ?????? . )  
 
 
function) 
(A) -
11
2
 
(B) -
7
2
 
(C) -6 
(D) -
17
2
 
Solution:  3[?? ] - 5
|?? |
?? = 3[?? ] - 5, if ?? > 0 
= 3[?? ] + 5, if ?? < 0 
? ?
-3/2
2
??? ( ?? ) ???? = ?
-3/2
-1
?( -1) ???? + ?
-1
0
?( 2) ???? + ?
0
1
?( -5) ???? + ?
1
2
?( -2) ???? 
= -1 (-1 +
3
2
) + 2( 1)+ 1( -5)+ ( -2)= -
1
2
+ 2 - 5 - 2 = -
11
2
 
Ans. (A) 
Problem3: The value of ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? , where [.] denotes the greatest integer function, is equal to - 
(A) 
5
4
+ v3 + (2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(B) 
5
4
+ v3 +
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(C) 
5
4
+
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(D) none of these 
Solution: We have,  ?? = ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? = ?
1
v2
?( ?? + 1) ???? + ?
v2
v3
?( ?? 2
+ 2
?? ) ???? + ?
v3
2
?( ?? 3
+ 3
?? ) ???? 
  = (
?? 2
2
+ ?? )
1
v2
+ (
?? 3
3
+
2
?? ?????? 2
)
v2
v3
+ (
?? 4
4
+
3
?? ?????? 3
)
v3
2
  
Problem4: Evaluate: ?
-10
20
?[?????? -1
 ?? ]???? . Here [.] is the greatest integer function. 
Solution:  ?? = ?
-10
20
?[?????? -1
 ?? ]???? , we know ?????? -1
 ?? ? ( 0, ?? ) ??? ? ?? 
Thus [?????? -1
 ?? ] = {3, ?? ? ( -8, ?????? 3) 2, ?? ? ( ?????? 3, ?????? 2) 1, ?? ? ( ?????? 2, ?????? 1) 0 ?? ? ( ?????? 1, 8)  
Hence ?? = ?
-10
?????? 3
?3???? + ?
?????? 3
?????? 2
?2???? + ?
?????? 2
?????? 1
?1???? + ?
?????? 1
20
?0???? = 30 + ?????? 1 + ?????? 2 + ?????? 3 
Ans. 
Do yourself -1: 
Evaluate: 
(i) ?
0
3
?|?? 2
- ?? - 2|???? 
(ii) ?
0
4
?{?? }???? , where {.} ?????????????? ???????????????????? ???????? ?? . 
(iii) ?
0
?? /2
?|?????? ?? - ?????? ?? |???? 
(iv) If ?? ( ?? ) = {2 0 = ?? = 1 ?? + [?? ] 1 = ?? < 3 , where [.] denotes the greatest integer function. Evaluate 
?
0
2
??? ( ?? ) ???? 
Page 4


DEFINITE INTEGRATION 
Imagine looking at the shadow cast by a tree at sunset. The length and shape of the shadow vary with 
the terrain and the tree's shape. To find the exact area of that shadow, we use the concept of the definite 
integral —a mathematical tool that calculates the area under a curve. 
A definite integral, written as ?
?? ?? ??? ( ?? ) ???? , measures the total area between a curve ?? ( ?? ) and the x-axis 
from ?? = ?? , ?? = ?? . This concept is crucial in fields like finance, where it helps quantify continuous 
change, track investments, and analyze economic trends, making it invaluable for informed decision-
making. 
1. THE FUNDAMENTAL THEOREM OF 
CALCULUS: 
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection 
between the two branches of calculus: differential calculus and integral calculus. Differential calculus 
arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, 
the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two 
problems are actually closely related. In fact, he realized that differentiation and integration are inverse 
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the 
derivative and the integral. It was Newton and Leibnitz who exploited this relationship and used it to 
develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental 
Theorem enabled them to compute areas and integrals very easily without having to compute them as 
limits of sums. 
 
The Fundamental Theorem of Calculus, Part 1 
If ?? is continuous on [?? , ?? ], then the function ?? defined by 
?? ( ?? ) = ? ?
?? ?? ??? ( ?? ) ???? ?? = ?? = ?? 
is continuous on [?? , ?? ] and differentiable on ( ?? , ?? ) , and ?? '
( ?? ) = ?? ( ?? ) . 
The Fundamental Theorem of Calculus, Part 2 
If ?? is continuous on [?? , ?? ], then 
? ?
?? ?? ??? ( ?? ) ???? = ?? ( ?? )- ?? ( ?? ) 
where ?? is any antiderivative of ?? , that is, a function such that ?? '
= ?? . 
Note: If ?
?? ?? ??? ( ?? ) ???? = 0 ? then the equation ?? ( ?? ) = 0 has atleast one root lying in ( ?? , ?? ) provided ?? is a 
continuous function in ( ?? , ?? ) . 
2. PROPERTIES OF DEFINITE INTEGRAL: 
Some properties which are quite useful while solving any kind of questions are: 
(a) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? provided ?? is same 
(b) ?
?? ?? ??? ( ?? ) ???? = -?
?? ?? ??? ( ?? ) ???? 
(c) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? + ?
?? ?? ??? ( ?? ) ???? , where c may lie inside or outside the interval [?? , ?? ]. This 
property is to be used when ?? is piecewise continuous in ( ?? , ?? ) . 
Problem1: If ?? ( ?? ) = {?? 2
, 0 < ?? < 2 3?? - 4,2 = ?? < 3  then evaluate ?
0
3
??? ( ?? ) ???? 
Solution:  ?
0
3
??? ( ?? ) ???? = ?
0
2
??? ( ?? ) ???? + ?
2
3
??? ( ?? ) ???? = ?
0
2
??? 2
???? + ?
2
3
?( 3?? - 4) ???? 
= (
?? 3
3
)
0
2
+ (
3?? 2
2
- 4?? )
2
3
=
8
3
+
27
2
- 12 - 6 + 8 = 37/6#( ?????? . )  
 
 
function) 
(A) -
11
2
 
(B) -
7
2
 
(C) -6 
(D) -
17
2
 
Solution:  3[?? ] - 5
|?? |
?? = 3[?? ] - 5, if ?? > 0 
= 3[?? ] + 5, if ?? < 0 
? ?
-3/2
2
??? ( ?? ) ???? = ?
-3/2
-1
?( -1) ???? + ?
-1
0
?( 2) ???? + ?
0
1
?( -5) ???? + ?
1
2
?( -2) ???? 
= -1 (-1 +
3
2
) + 2( 1)+ 1( -5)+ ( -2)= -
1
2
+ 2 - 5 - 2 = -
11
2
 
Ans. (A) 
Problem3: The value of ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? , where [.] denotes the greatest integer function, is equal to - 
(A) 
5
4
+ v3 + (2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(B) 
5
4
+ v3 +
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(C) 
5
4
+
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(D) none of these 
Solution: We have,  ?? = ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? = ?
1
v2
?( ?? + 1) ???? + ?
v2
v3
?( ?? 2
+ 2
?? ) ???? + ?
v3
2
?( ?? 3
+ 3
?? ) ???? 
  = (
?? 2
2
+ ?? )
1
v2
+ (
?? 3
3
+
2
?? ?????? 2
)
v2
v3
+ (
?? 4
4
+
3
?? ?????? 3
)
v3
2
  
Problem4: Evaluate: ?
-10
20
?[?????? -1
 ?? ]???? . Here [.] is the greatest integer function. 
Solution:  ?? = ?
-10
20
?[?????? -1
 ?? ]???? , we know ?????? -1
 ?? ? ( 0, ?? ) ??? ? ?? 
Thus [?????? -1
 ?? ] = {3, ?? ? ( -8, ?????? 3) 2, ?? ? ( ?????? 3, ?????? 2) 1, ?? ? ( ?????? 2, ?????? 1) 0 ?? ? ( ?????? 1, 8)  
Hence ?? = ?
-10
?????? 3
?3???? + ?
?????? 3
?????? 2
?2???? + ?
?????? 2
?????? 1
?1???? + ?
?????? 1
20
?0???? = 30 + ?????? 1 + ?????? 2 + ?????? 3 
Ans. 
Do yourself -1: 
Evaluate: 
(i) ?
0
3
?|?? 2
- ?? - 2|???? 
(ii) ?
0
4
?{?? }???? , where {.} ?????????????? ???????????????????? ???????? ?? . 
(iii) ?
0
?? /2
?|?????? ?? - ?????? ?? |???? 
(iv) If ?? ( ?? ) = {2 0 = ?? = 1 ?? + [?? ] 1 = ?? < 3 , where [.] denotes the greatest integer function. Evaluate 
?
0
2
??? ( ?? ) ???? 
(d) ?
-?? ?? ??? ( ?? ) ???? = ?
0
?? ?[?? ( ?? )+ ?? ( -?? ) ]???? =
[0 ; ???? ?? ( ?? ) ???? ???? ?????? ????????????????  2?
0
?? ??? ( ?? ) ???? ; ???? ?? ( ?? ) ???? ???? ???????? ????????????????   
Problem5: Evaluate ?
-1/2
1/2
??????? ?????? (
1+?? 1-?? )???? 
Solution:  ?? ( -?? ) = ?????? ( -?? ) ???? (
1-?? 1+?? ) = -?????? ???? (
1+?? 1-?? ) = -?? ( ?? ) 
? ?? ( ?? ) is odd 
Hence, the value of the given integral = 0. 
Problem6: If ?? ( ?? ) = |?????? ?? ?? ?? 2
 2?? ?????? 2
 ?? /2 ?? 2
 ?????? ?? ?????? ?? + ?? 3
 1 2 ?? + ?????? ?? |, then the value of 
?
-?? /2
?? /2
?( ?? 2
+ 1) ( ?? ( ?? )+ ?? ''
( ?? ) ) ???? 
(A) 1 
(B) -1 
(C) 2 
(D) none of these 
Solution:   As, ?? ( ?? ) = |?????? ?? ?? ?? 2
 2?? ?????? 2
 ?? /2 ?? 2
 ?????? ?? ?????? ?? + ?? 3
 1 2 ?? + ?????? ?? | 
  ? ?? ( -?? ) = -?? ( ?? ) ? ?? ( ?? ) ???? ??????    ? ?? '
( ?? ) ???? ????????  ? ?? ''
( ?? ) ???? ??????   
Thus, ?? ( ?? )+ ?? ''
( ?? ) is odd function let, 
 ?? ( ?? )= ( ?? 2
+ 1)· {?? ( ?? )+ ?? ''
( ?? ) } ?  ?? ( -?? ) = -?? ( ?? )  
i.e. ?? ( ?? ) is odd 
? ? ?
?? /2
-?? /2
???? ( ?? ) ???? = 0#( ?? )  
Problem7: If ?? , ?? , h be continuous functions on [0, ?? ] such that ?? ( ?? - ?? ) = -?? ( ?? ) , ?? ( ?? - ?? ) = ?? ( ?? ) and 
3h( ?? )- 4h( ?? - ?? )= 5, then prove that ?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? ) ???? = 0 
Solution:  ?? = ?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? ) ???? = ?
0
?? ??? ( ?? - ?? ) ?? ( ?? - ?? ) h( ?? - ?? ) ???? = -?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? - ?? ) ???? 7?? =
3?? + 4?? = ?
0
?? ??? ( ?? ) ?? ( ?? ) {3h( ?? )- 4h( ?? - ?? ) }???? = 5?
0
?? ??? ( ?? ) ?? ( ?? ) ???? = 0 
(since ?? ( ?? - ?? ) ?? ( ?? - ?? ) = -?? ( ?? ) ?? ( ?? ) ) 
? ?? = 0 
Problem8:   Evaluate ?
-?? ?? ?
???????? ?? ?? ?? +1
???? 
Solution:  ?? = ?
-?? 0
?
???????? ?? ?? ?? +1
???? + ?
0
?? ?
???????? ?? ?? ?? +1
???? = ?? 1
+ ?? 2
 
where ?? 1
= ?
-?? 0
?
???????? ?? ?? ?? +1
???? 
Page 5


DEFINITE INTEGRATION 
Imagine looking at the shadow cast by a tree at sunset. The length and shape of the shadow vary with 
the terrain and the tree's shape. To find the exact area of that shadow, we use the concept of the definite 
integral —a mathematical tool that calculates the area under a curve. 
A definite integral, written as ?
?? ?? ??? ( ?? ) ???? , measures the total area between a curve ?? ( ?? ) and the x-axis 
from ?? = ?? , ?? = ?? . This concept is crucial in fields like finance, where it helps quantify continuous 
change, track investments, and analyze economic trends, making it invaluable for informed decision-
making. 
1. THE FUNDAMENTAL THEOREM OF 
CALCULUS: 
The Fundamental Theorem of Calculus is appropriately named because it establishes a connection 
between the two branches of calculus: differential calculus and integral calculus. Differential calculus 
arose from the tangent problem, whereas integral calculus arose from a seemingly unrelated problem, 
the area problem. Newton's teacher at Cambridge, Isaac Barrow (1630-1677), discovered that these two 
problems are actually closely related. In fact, he realized that differentiation and integration are inverse 
processes. The Fundamental Theorem of Calculus gives the precise inverse relationship between the 
derivative and the integral. It was Newton and Leibnitz who exploited this relationship and used it to 
develop calculus into a systematic mathematical method. In particular, they saw that the Fundamental 
Theorem enabled them to compute areas and integrals very easily without having to compute them as 
limits of sums. 
 
The Fundamental Theorem of Calculus, Part 1 
If ?? is continuous on [?? , ?? ], then the function ?? defined by 
?? ( ?? ) = ? ?
?? ?? ??? ( ?? ) ???? ?? = ?? = ?? 
is continuous on [?? , ?? ] and differentiable on ( ?? , ?? ) , and ?? '
( ?? ) = ?? ( ?? ) . 
The Fundamental Theorem of Calculus, Part 2 
If ?? is continuous on [?? , ?? ], then 
? ?
?? ?? ??? ( ?? ) ???? = ?? ( ?? )- ?? ( ?? ) 
where ?? is any antiderivative of ?? , that is, a function such that ?? '
= ?? . 
Note: If ?
?? ?? ??? ( ?? ) ???? = 0 ? then the equation ?? ( ?? ) = 0 has atleast one root lying in ( ?? , ?? ) provided ?? is a 
continuous function in ( ?? , ?? ) . 
2. PROPERTIES OF DEFINITE INTEGRAL: 
Some properties which are quite useful while solving any kind of questions are: 
(a) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? provided ?? is same 
(b) ?
?? ?? ??? ( ?? ) ???? = -?
?? ?? ??? ( ?? ) ???? 
(c) ?
?? ?? ??? ( ?? ) ???? = ?
?? ?? ??? ( ?? ) ???? + ?
?? ?? ??? ( ?? ) ???? , where c may lie inside or outside the interval [?? , ?? ]. This 
property is to be used when ?? is piecewise continuous in ( ?? , ?? ) . 
Problem1: If ?? ( ?? ) = {?? 2
, 0 < ?? < 2 3?? - 4,2 = ?? < 3  then evaluate ?
0
3
??? ( ?? ) ???? 
Solution:  ?
0
3
??? ( ?? ) ???? = ?
0
2
??? ( ?? ) ???? + ?
2
3
??? ( ?? ) ???? = ?
0
2
??? 2
???? + ?
2
3
?( 3?? - 4) ???? 
= (
?? 3
3
)
0
2
+ (
3?? 2
2
- 4?? )
2
3
=
8
3
+
27
2
- 12 - 6 + 8 = 37/6#( ?????? . )  
 
 
function) 
(A) -
11
2
 
(B) -
7
2
 
(C) -6 
(D) -
17
2
 
Solution:  3[?? ] - 5
|?? |
?? = 3[?? ] - 5, if ?? > 0 
= 3[?? ] + 5, if ?? < 0 
? ?
-3/2
2
??? ( ?? ) ???? = ?
-3/2
-1
?( -1) ???? + ?
-1
0
?( 2) ???? + ?
0
1
?( -5) ???? + ?
1
2
?( -2) ???? 
= -1 (-1 +
3
2
) + 2( 1)+ 1( -5)+ ( -2)= -
1
2
+ 2 - 5 - 2 = -
11
2
 
Ans. (A) 
Problem3: The value of ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? , where [.] denotes the greatest integer function, is equal to - 
(A) 
5
4
+ v3 + (2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(B) 
5
4
+ v3 +
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(C) 
5
4
+
v2
3
+
1
?????? 2
(2
v3
- 2
v2
) +
1
?????? 3
(9 - 3
v3
) 
(D) none of these 
Solution: We have,  ?? = ?
1
2
?( ?? [?? 2
]
+ [?? 2
]
?? ) ???? = ?
1
v2
?( ?? + 1) ???? + ?
v2
v3
?( ?? 2
+ 2
?? ) ???? + ?
v3
2
?( ?? 3
+ 3
?? ) ???? 
  = (
?? 2
2
+ ?? )
1
v2
+ (
?? 3
3
+
2
?? ?????? 2
)
v2
v3
+ (
?? 4
4
+
3
?? ?????? 3
)
v3
2
  
Problem4: Evaluate: ?
-10
20
?[?????? -1
 ?? ]???? . Here [.] is the greatest integer function. 
Solution:  ?? = ?
-10
20
?[?????? -1
 ?? ]???? , we know ?????? -1
 ?? ? ( 0, ?? ) ??? ? ?? 
Thus [?????? -1
 ?? ] = {3, ?? ? ( -8, ?????? 3) 2, ?? ? ( ?????? 3, ?????? 2) 1, ?? ? ( ?????? 2, ?????? 1) 0 ?? ? ( ?????? 1, 8)  
Hence ?? = ?
-10
?????? 3
?3???? + ?
?????? 3
?????? 2
?2???? + ?
?????? 2
?????? 1
?1???? + ?
?????? 1
20
?0???? = 30 + ?????? 1 + ?????? 2 + ?????? 3 
Ans. 
Do yourself -1: 
Evaluate: 
(i) ?
0
3
?|?? 2
- ?? - 2|???? 
(ii) ?
0
4
?{?? }???? , where {.} ?????????????? ???????????????????? ???????? ?? . 
(iii) ?
0
?? /2
?|?????? ?? - ?????? ?? |???? 
(iv) If ?? ( ?? ) = {2 0 = ?? = 1 ?? + [?? ] 1 = ?? < 3 , where [.] denotes the greatest integer function. Evaluate 
?
0
2
??? ( ?? ) ???? 
(d) ?
-?? ?? ??? ( ?? ) ???? = ?
0
?? ?[?? ( ?? )+ ?? ( -?? ) ]???? =
[0 ; ???? ?? ( ?? ) ???? ???? ?????? ????????????????  2?
0
?? ??? ( ?? ) ???? ; ???? ?? ( ?? ) ???? ???? ???????? ????????????????   
Problem5: Evaluate ?
-1/2
1/2
??????? ?????? (
1+?? 1-?? )???? 
Solution:  ?? ( -?? ) = ?????? ( -?? ) ???? (
1-?? 1+?? ) = -?????? ???? (
1+?? 1-?? ) = -?? ( ?? ) 
? ?? ( ?? ) is odd 
Hence, the value of the given integral = 0. 
Problem6: If ?? ( ?? ) = |?????? ?? ?? ?? 2
 2?? ?????? 2
 ?? /2 ?? 2
 ?????? ?? ?????? ?? + ?? 3
 1 2 ?? + ?????? ?? |, then the value of 
?
-?? /2
?? /2
?( ?? 2
+ 1) ( ?? ( ?? )+ ?? ''
( ?? ) ) ???? 
(A) 1 
(B) -1 
(C) 2 
(D) none of these 
Solution:   As, ?? ( ?? ) = |?????? ?? ?? ?? 2
 2?? ?????? 2
 ?? /2 ?? 2
 ?????? ?? ?????? ?? + ?? 3
 1 2 ?? + ?????? ?? | 
  ? ?? ( -?? ) = -?? ( ?? ) ? ?? ( ?? ) ???? ??????    ? ?? '
( ?? ) ???? ????????  ? ?? ''
( ?? ) ???? ??????   
Thus, ?? ( ?? )+ ?? ''
( ?? ) is odd function let, 
 ?? ( ?? )= ( ?? 2
+ 1)· {?? ( ?? )+ ?? ''
( ?? ) } ?  ?? ( -?? ) = -?? ( ?? )  
i.e. ?? ( ?? ) is odd 
? ? ?
?? /2
-?? /2
???? ( ?? ) ???? = 0#( ?? )  
Problem7: If ?? , ?? , h be continuous functions on [0, ?? ] such that ?? ( ?? - ?? ) = -?? ( ?? ) , ?? ( ?? - ?? ) = ?? ( ?? ) and 
3h( ?? )- 4h( ?? - ?? )= 5, then prove that ?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? ) ???? = 0 
Solution:  ?? = ?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? ) ???? = ?
0
?? ??? ( ?? - ?? ) ?? ( ?? - ?? ) h( ?? - ?? ) ???? = -?
0
?? ??? ( ?? ) ?? ( ?? ) h( ?? - ?? ) ???? 7?? =
3?? + 4?? = ?
0
?? ??? ( ?? ) ?? ( ?? ) {3h( ?? )- 4h( ?? - ?? ) }???? = 5?
0
?? ??? ( ?? ) ?? ( ?? ) ???? = 0 
(since ?? ( ?? - ?? ) ?? ( ?? - ?? ) = -?? ( ?? ) ?? ( ?? ) ) 
? ?? = 0 
Problem8:   Evaluate ?
-?? ?? ?
???????? ?? ?? ?? +1
???? 
Solution:  ?? = ?
-?? 0
?
???????? ?? ?? ?? +1
???? + ?
0
?? ?
???????? ?? ?? ?? +1
???? = ?? 1
+ ?? 2
 
where ?? 1
= ?
-?? 0
?
???????? ?? ?? ?? +1
???? 
Put ?? = -?? ? ???? = -???? 
? ?? 1
= ?
?? 0
?
( -?? ) ?????? ( -?? ) ( -???? )
?? -?? + 1
= ?
0
?? ?
???????? ?????? ?? -?? + 1
= ?
0
?? ?
?? ?? ???????? ?????? ?? ?? + 1
= ?
0
?? ?
?? ?? ???????? ?????? ?? ?? + 1
 
Hence ?? = ?? 1
+ ?? 2
= ?
0
?? ?
?? ?? ???????? ?? ?? ?? +1
???? + ?
0
?? ?
???????? ?? ?? ?? +1
???? 
?? = ?
0
?? ????????? ?????? = ?
0
?? ?( ?? - ?? ) ?????? ( ?? - ?? ) ???? = ?? ?
0
?? ??????? ?????? - ?? 
? 2?? = ?? ?
0
?? ??????? ?????? = ?? | - ?????? ?? |
0
?? = 2?? ? ?? = ?? 
Problem9: Evaluate ?
0
2
?
????
( 17+8?? -4?? 2
) [?? 6( 1-?? )
+1]
 
Solution:   Let ?? = ?
0
2
?
????
( 17+8?? -4?? 2
) [?? 6( 1-?? )
+1]
 
Also ?? = ?
0
2
?
????
( 17+8?? -4?? 2
) [?? -6( 1-?? )
+1]
[? ?
0
?? ??? ( ?? ) ???? = ?
0
?? ??? ( ?? - ?? ) ???? ] 
Adding, we get 
2??  = ? ?
2
0
??
1
17 + 8?? - 4?? 2
(
1
?? 6( 1-?? )
+ 1
+
1
?? -6( 1-?? )
+ 1
)????   = ? ?
2
0
??
1
17 + 8?? - 4?? 2
????
= -
1
4
? ?
2
0
??
????
?? 2
- 2?? - 17/4
   = -
1
4
? ?
2
0
??
????
( ?? - 1)
2
- 21/4
= -
1
4
×
1
2 ×
v 21
2
[?????? |
?? - 1 -
v 21
2
?? - 1 +
v 21
2
|]
0
2
   = -
1
4v 21
[?????? |
2?? - 2 - v 21
2?? - 2 + v 21
|]
0
2
? ?? = -
1
8v 21
[?????? |
2 - v 21
2 + v 21
| - ?????? |
2 + v 21
v 21 - 2
|]   = -
1
4v 21
[?????? |
v 21 - 2
2 + v 21
|]  
Problem10:  ?
0
1
??????? -1
 ( 1 - ?? + ?? 2
) ???? equals - 
(A) 
?? 2
+ ?????? 2 
(B) 
?? 2
- ?????? 2 
(C) ?? - ?????? 2 
(D) none of these 
Solution:  ?? = ?
0
1
??????? -1
 (
1
1-?? +?? 2
)???? = ?
0
1
??????? -1
 (
?? +( 1-?? )
1-?? ( 1-?? )
)???? 
  = ? ?
1
0
??[?????? -1
 ?? + ?????? -1
 ( 1 - ?? ) ]???? = ? ?
1
0
???????? -1
 ?????? + ? ?
1
0
???????? -1
 ( 1 - ?? ) ????  
Problem11: ?
0
?? /2
?
???????? ?? +???????? ?? ?????? ?? +?????? ?? ???? 
Solution:  ?? = ?
0
?? /2
?
???????? ?? +???????? ?? ?????? ?? +?????? ?? ????