Properties
of Definite Integral
- Change
of variable of integration is immaterial so long as limits of integration
remain the same i.e.
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image001.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image002.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image003.gif)
Where
the point c may lie between a and b or it may be exterior to (a, b). .
Note:
This property is useful when f (x) is not continuous in
[a, b] because we can break up the integral into several integrals at the
points of discontinuity so that the function is continuous in the subintervals.
Example.1
Evaluate
.
Solution
Let
.
Then
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image006.gif)
Hence, I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image007.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image008.gif)
=
=
2+1+6 = 9
Note:
The expression for
f(x) changes at the end points of each of the sub-interval. You might at times
be puzzled about the end points as limits of integration. You may not be sure
for x=0 as the limit of the 1st integral or the next one. In fact, it is immaterial,
as the area of the line is always zero. Therefore even if you write
, whereas 0 is not
included in its domain it is deemed to be understood that you are approaching x
= 0 from the left in the first integral and from the right in the second
integral. Similarly
for x = 1.
Example.2
Evaluate
.
Solution
Put x =
+ t
&⇒ dx = dt
I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image013.gif)
I =
(
|sin t| has period p)
I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image016.gif)
4. ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image017.gif)
In particular
.
Example.1
Evaluate
.
Solution
I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image020.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image021.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image022.gif)
&⇒ I = – I
&⇒ 2I = 0
&⇒ I = 0
Example.2
Evaluate
.
Solution
I
=
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image025.gif)
=
(since a + b = p/2)
Hence, 2I =
= p/3 – p/6 = p/6
&⇒ I = p/12
5.![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image028.gif)
Special
cases:
If f (x) = f (a – x), then
.
If
f (x) = - f (a – x), then
.
Example.1
Evaluate
.
Solution
Let
I =
=
(by property 4 )
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image034.gif)
&⇒
2 I =
=
(by property 5)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image037.gif)
Let tan x = t so that for x → 0, t → 0 and for x →
p/2, t → ¥.
Hence we can write, I =
=
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image039.gif)
=
=
=
.
6.![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image043.gif)
Special
case:
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image044.gif)
Example.1
Evaluate
.
Solution
f(-x)
= cos (-x) ln
= - cosx ln
= - f(x)
&⇒
f(x) is odd.
Hence,
the value of the given integral = 0
Example.2
Prove that
.
Solution
Let
so that dx = dt and
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image050.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image051.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image052.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image053.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image055.gif)
Alternative:
I =
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image057.gif)
&⇒ 2I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image058.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image059.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image060.gif)
Let x -
= t
2I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image059.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image063.gif)
2![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image059.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image064.gif)
I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image059.gif)
.
7. ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image065.gif)
Example.1
Evaluate
.
Solution
Let I1
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image067.gif)
=
=
.
Let I2 =
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image071.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image072.gif)
&⇒ I1 + 3 I2 = 0
8. If f (x) is a periodic function with period T, then.
(a)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image073.gif)
(i) In particular, if a = 0
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image074.gif)
(ii) If n = 1
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image075.gif)
(b)![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image076.gif)
(c)![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image077.gif)
Example.1
Evaluate
.
Solution
Note that |cos x| is a
periodic function with period p. Hence the given integral .
I =
(using
property 8)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image080.gif)
=
=
4 [1 + 1] = 8
Example.2
Show
that
, where n is a positive
integer and 0 £ v < p
Solution
![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image083.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image085.gif)
=
– cos v + 1 + n ![](http://www.quizsolver.com/radix/dth/notif/Definite%20integral%203_files/image086.gif)
=
– cos v + 1 + n (1 + 1) = 2n + 1 – cos v