Average value by integral method

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1 INTEGRAL AVERAGE VALUE BYINTEGRAL METHOD

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

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3 DRAFT COPY -GPL LICENSING

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4 Average Value We knowthat 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:

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5 1. Whether thequantity, 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

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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.

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7 π 3 2π 3 π 4π 3 5π 32π 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|

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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)

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9 Solved Problem 0.1Find 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

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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

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11 t f(t) b b f(a) f(b) b b a b tline 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.

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12 Solution From theaverage 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

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13 Substituting the valuesand 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.

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14 Solution The averageof 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)

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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

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16 1 −1 1 2 34 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.

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17 1 −1 1 2 34 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

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18 The function isabsolute additive, hence Avg =

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1 R 0 f(x) dx

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2 R 1 f(x) dx

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2 R 0 dx Or Avg =

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1 R 0 1 dx

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2 R 1 ‘ − 1dx

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2 R 0 dx On integrating andsimplifying, we have Avg = 1 + 1 2 = 1

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19 Thus average valueof 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.

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20 Solved Problem 0.6Find 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.

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21 1 −1 1 2 34 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

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1 R 0 f(x) dx

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3 R 1 f(x) dx

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3 R 0 dx Or Avg =

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1 R 0 1 dx

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3 R 1 −1 dx

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3 R 0 dx On integrating andsimplifying, we have Avg = 1 + 2 3 = 1

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23 Thus average valueof 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.

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24 Solved Problem 0.7Find 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.

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1 R 0 f′ (x) dx

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2 R 1 |f′′ (x) dx

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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

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27 0.5 1.0 −0.5 −1.0 0.5 1.0 1.52.0 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

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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)

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

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1 R 0 x dx

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+

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2 R 1 −(2 − x)dx

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2 R 0 dx Or Avg =

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x2 2

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1 0 +

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2x − x2 2

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2 1 2 On substituting thelimits, we have Avg = 1 2 + 1 2 2 On simplifying, we have Avg = 1 2

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30 Thus average valueof 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

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31 On integrating andsimplifying, 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.

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1 R 0 f′ (x) dx

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+

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2 R 1 |f′′ (x) dx

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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

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34 0.5 1.0 −0.5 −1.0 0.5 1.0 1.52.0 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

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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)

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

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1 R 0 0.5(1 + x)dx

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+

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2 R 1 −(1.5 − 0.5x)dx

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2 R 0 dx Or Avg =

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0.5x + 0.5x2 2

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1 0 +

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1.5x − 0.5x2 2

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2 1 2 On substituting thelimits, we have Avg = 0.75 + 0.75 2 = 0.75

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37 Thus average valueof the function is three quarter. 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 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

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

Average value by integral method

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Average value by integral method