Average of continuous or discrete function by using integral method. For K12 Students.

1. 1
INTEGRAL
AVERAGE VALUE BY INTEGRAL METHOD

2. 2
Arun Umrao
https://sites.google.com/view/arunumrao

3. 3
DRAFT COPY - GPL LICENSING

4. 4
Average Value
We know that
Z
f(x) dx represents area between function f(x) and two points of
width dx in base line. If
b
Z
a
f(x) dx
is area between f(x) and x-limits (between x = a and x = b) then average value for
this limit is given by
Iavg =
b
R
a
f(x) dx
b
R
a
dx
(1)
Above relation is similar to the mathematical average, i.e. average area is the ratio of
total area to the difference of limit. While we are using this average relation, following
points must be keep in mind:

5. 5
1. Whether the quantity, whose average value is being evaluated, has zero crossing
or not within the given range of limits. In other words, is quantity is zero at
any point within the limits? For example, sine function has zero crossing at π
and its multiples.
π
3
2π
3 π 4π
3
5π
3 2π 7π
3
8π
3 3π
x
y
b
b
b
b
b b b
b
b
b
b
b
b
b
b b b
b
b
b
b
b
b
b
b b b
b
b
b
b
b
b
sin x
Zero Crossing
If there is zero crossing, function is solved by using piecewise method. For
example, if f(x) is zero at c (in limits a < c < b) then average of function will

6. 6
be computed as
Iavg =
1
b
R
a
dx


c
Z
a
f(x) dx +
b
Z
c
f′
(x) dx


Where, f(x) and f′
(x) are effective function1
before and after zero crossing.
2. The quantity whose average value is being evaluated is periodic or not. If its is
periodic then it replicates itself after fixed interval and we can find the average
value by multiplying number of replications with average value of one period.
For example, for sine function is periodic function of 2π interval.
1
Functions are made absolute or not it depends what the function is representing.

7. 7
π
3
2π
3 π 4π
3
5π
3 2π 7π
3
8π
3 3π
x
y
b
b
b
b
b b b
b
b
b
b
b
b
b
b b b
b
b
b
b
b
b
b
b b b
b
b
b
b
b
b
sin x
One Period
3. Whether the physical quantity is absolute additive or algebraic additive. For
example, area is absolute additive. It is never negative. So, even if mathematical
computations are negative, they are added rather than subtraction.
π
3
2π
3 π 4π
3
5π
3 2π 7π
3
8π
3 3π
x
y
sin x
|A|
| − A|

8. 8
Therefore,
A = |A| + | − A|
Similarly, profit is algebraically additive. Profit of positive side is get subtracted
by loss in negative side. In other-words, positive profit is subtracted by negative
profit.
42 84 126 168 210 252 294 336 378
n
U
Cost Function
P
−P
Therefore,
P = P + (−P)

9. 9
Solved Problem 0.1 Find the average value of constant ‘c’ for the limits ranges from
t = a to t = b.
Solution The average value of function f(t) is given by
Iavg =
b
R
a
c dt
b
R
a
dt
t
c b b
c c
b b
a b
Now
Iavg = c
[t]
b
a
[t]
b
a

10. 10
Solving the relation
Iavg = c
This shows that the average value of constant remains constant.
Solved Problem 0.2 Find the average value of f(t) = t2
for the t-limits ranges from
t = a to t = b.
Solution The average value of function f(t) is given by
Iavg =
b
R
a
t dt
b
R
a
dt

11. 11
t
f(t)
b
b
f(a)
f(b)
b b
a b
t line
Now
Iavg =
h
t2
2
ib
a
[t]
b
a
Solving the relation
Iavg =
1
2
(a + b)
This shows that the average values of a line is average of sum of two ends of the line.
Solved Problem 0.3 Find the average value of f(x) = x2
for the x-limits ranges from
x = 2 to x = 3.

12. 12
Solution From the average relation
Iavg =
b
R
a
f(x) dx
b
R
a
dx
x
f(x)
b
b
f(a)
f(b)
b b
a b
f(x) = x2

13. 13
Substituting the values and integrating for the limits
Iavg =
3
R
2
x2
dx
3
R
2
dx
Or
Iavg =
h
x3
3
i3
2
[x]
3
2
Or
Iavg =
19
3
This is the required average value.
Solved Problem 0.4 Find the average value of f(x) = sin(x) within the angle limits
in radian values from x = π/3 to 2π/3.

14. 14
Solution The average of function within the limits is
Iavg =
2π
3
R
π
6
sin(x)dx
2π
3
R
π
6
dx
On integration and solving it
Iavg =
[− cos(x)]
2π
3
π
6
[x]
2π
3
π
6
x
f(x)
b
b
f(π/6)
f(2π/3)
b b
π/6 2π/3
f(x) = sin(x)

15. 15
Iavg =
− cos(2π
3 ) + cos(π
6 )
2π
3 − π
6
Or
Iavg =
0.5 + 0.866)
2.093 − 0.523
=
1.366
1.57
= 0.87
This is required result.
Solved Problem 0.5 Find the periodic average value of the following function.
Solution

16. 16
1
−1
1 2 3 4 5
x
f(x)
This function is a periodic function of period 2 and has equal width in both half
cycles. So, average value of one cycle of period 2 is equal to the average value of other
cycles. There are two cases for average values.

17. 17
1
−1
1 2 3 4 5
x
f(x)
If function is absolute additive then average value of the function is
A =
2
R
0
f(x) dx
2
R
0
dx

18. 18
The function is absolute additive, hence
Avg =

22. 1
R
0
f(x) dx

26. +

30. 2
R
1
f(x) dx

34. 2
R
0
dx
Or
Avg =

38. 1
R
0
1 dx

42. +

46. 2
R
1
‘ − 1 dx

50. 2
R
0
dx
On integrating and simplifying, we have
Avg =
1 + 1
2
= 1

51. 19
Thus average value of the function is one. If function is algebraically additive, then
Avg =
1
R
0
f(x) dx +
2
R
1
f(x) dx
2
R
0
dx
Or
Avg =
1
R
0
1 dx +
2
R
1
−1 dx
2
R
0
dx
On integrating and simplifying, we have
Avg =
1 − 1
2
= 0
In this case, average value is zero.

52. 20
Solved Problem 0.6 Find the periodic average value of the following function.
Solution
1
−1
1 2 3 4 5 6
x
f(x)
This function is a periodic function of period 3 and has unequal width in both
half cycles. So, average value of one cycle of period 2 is equal to the average value of
other cycles. There are two cases for average values.

53. 21
1
−1
1 2 3 4 5 6
x
f(x)
If function is absolute additive then average value of the function is
A =
3
R
0
f(x) dx
3
R
0
dx

54. 22
The function is absolute additive, hence
Avg =

58. 1
R
0
f(x) dx

62. +

66. 3
R
1
f(x) dx

70. 3
R
0
dx
Or
Avg =

74. 1
R
0
1 dx

78. +

82. 3
R
1
−1 dx

86. 3
R
0
dx
On integrating and simplifying, we have
Avg =
1 + 2
3
= 1

87. 23
Thus average value of the function is one. If function is algebraically additive, then
Avg =
1
R
0
f(x) dx +
3
R
1
f(x) dx
3
R
0
dx
Or
Avg =
1
R
0
1 dx +
3
R
1
−1 dx
3
R
0
dx
On integrating and simplifying, we have
Avg =
1 − 2
3
= −
1
3
In this case, average value is −1/3.

88. 24
Solved Problem 0.7 Find the periodic average value of the following function.
Solution
1
−1
1 2 3 4 5 6
x
f(x)
This is periodic function of period 2. Is shape is like saw tooth.

89. 25
1
−1
1 2 3 4 5 6
x
f(x)
If function is absolute additive then average value of the function is
A =
2
R
0
f(x) dx
2
R
0
dx

90. 26
The function is absolute additive, hence
Avg =

94. 1
R
0
f′
(x) dx

98. +

102. 2
R
1
|f′′
(x) dx

106. 2
R
0
dx
Before moving ahead, we will construct the function of both half cycles. For this
purpose both half cycles of one cycle of the function is re-graphed as

107. 27
0.5
1.0
−0.5
−1.0
0.5 1.0 1.5 2.0
x
f(x)
y1
x1
y2
x2
For region 0 to 1, function is line segment. The slope of line segments (0, 0) to
(x1, y1) and (0, 0) to (1, 1) are equal. So
y1 − 0
x1 − 0
=
1 − 0
1 − 0

108. 28
It gives,
y1 = x1
Transform this relation into function form and we get first half cycle as function
f′
(x) = x. Similarly, for second half cycle, we have
0 − (y2)
2 − x2
=
0 − (−1)
2 − 1
Note that, here we have not taken sign with y2 as it is dependent variable (or say
function symbol) and in algebraic equations, dependent variable (function symbol)
is always positive and sign in function result is determined by independent variable
terms. This denomination is also called standardisation of algebraic equation. Now,
above equation gives,
y2 = −(2 − x2)

109. 29
Transform this relation into function form and we get second half cycle as function
f′′
(x) = −(2 − x). Now, average value is
Avg =

113. 1
R
0
x dx

117. +

121. 2
R
1
−(2 − x) dx

125. 2
R
0
dx
Or
Avg =

128. x2
2

131. 1
0
+

134. 2x − x2
2

137. 2
1
2
On substituting the limits, we have
Avg =
1
2 + 1
2
2
On simplifying, we have
Avg =
1
2

138. 30
Thus average value of the function is half. If function is algebraically additive, then
Avg =
1
R
0
f′
(x) dx +
2
R
1
f′′
(x) dx
2
R
0
dx
Or
Avg =
1
R
0
x dx +
3
R
1
−(2 − x) dx
2
R
0
dx
Or
Avg =
1
R
0
x dx −
3
R
1
(2 − x) dx
2
R
0
dx

139. 31
On integrating and simplifying, we have
Avg =
1
2 − 1
2
2
= 0
In this case, average value is zero.
Solved Problem 0.8 Find the periodic average value of the following function.
Solution
1
−1
1 2 3 4 5 6
x
f(x)
This is periodic function of period 2. Is shape is like saw tooth.

140. 32
1
−1
1 2 3 4 5 6
x
f(x)
If function is absolute additive then average value of the function is
A =
2
R
0
f(x) dx
2
R
0
dx

141. 33
The function is absolute additive, hence
Avg =

145. 1
R
0
f′
(x) dx

149. +

153. 2
R
1
|f′′
(x) dx

157. 2
R
0
dx
Before moving ahead, we will construct the function of both half cycles. For this
purpose both half cycles of one cycle of the function is re-graphed as

158. 34
0.5
1.0
−0.5
−1.0
0.5 1.0 1.5 2.0
x
f(x)
y1
x1
y2
x2
For region 0 to 1, function is line segment. The slope of line segments (0, 0.5) to
(x1, y1) and (0, 0.5) to (1, 1) are equal. So
y1 − 0.5
x1 − 0
=
1 − 0.5
1 − 0

159. 35
It gives,
y1 = 0.5(1 + x1)
Transform this relation into function form and we get first half cycle as function
f′
(x) = 0.5(1 + x). Similarly, for second half cycle, we have
−0.5 − (y2)
2 − x2
=
−0.5 − (−1)
2 − 1
Note that, here we have not taken sign with y2 as it is dependent variable (or say
function symbol) and in algebraic equations, dependent variable (function symbol)
is always positive and sign in function result is determined by independent variable
terms. This denomination is also called standardisation of algebraic equation. Now,
above equation gives,
y2 = −(1.5 − 0.5x2)

160. 36
Transform this relation into function form and we get second half cycle as function
f′′
(x) = −(1.5 − 0.5x). Now, average value is
Avg =

164. 1
R
0
0.5(1 + x) dx

168. +

172. 2
R
1
−(1.5 − 0.5x) dx

176. 2
R
0
dx
Or
Avg =

179. 0.5x + 0.5x2
2

182. 1
0
+

185. 1.5x − 0.5x2
2

188. 2
1
2
On substituting the limits, we have
Avg =
0.75 + 0.75
2
= 0.75

189. 37
Thus average value of the function is three quarter. If function is algebraically addi-
tive, then
Avg =
1
R
0
f′
(x) dx +
2
R
1
f′′
(x) dx
2
R
0
dx
Or
Avg =
1
R
0
0.5(1 + x) dx +
3
R
1
−(1.5 − 0.5x) dx
2
R
0
dx
Or
Avg =
1
R
0
0.5(1 + x) dx −
3
R
1
(1.5 − 0.5x) dx
2
R
0
dx

190. 38
On integrating and simplifying, we have
Avg =
0.75 − 0.75
2
= 0
In this case, average value is zero.