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Average Value Of Functions

Find the average value of the function over an interval & learn average value formula & calculus. Practice mean value theorem calculus problems for Integrals.

Average Value of Given Function

• If we are given finitely many numbers say $$x_1, x_2,x_3,...,x_n$$, then the average value of these values is given by

$$x_{(avg)}=\dfrac {x_1+x_2+x_3+...+x_n}{n}$$

(where n is the number of numbers)

$$=\dfrac {\sum\limits_{i=1}^{n}\;x_i}{n}$$

Average value of a function

• Suppose a function  $$y = f(x)$$  is defined in a closed interval $$[a, b]$$ and the average value of the function is desired in this interval, since there are infinitely many values of  $$x$$ in any interval,  the formula  $$\dfrac{\sum x_i}{n}$$ will not work.

We divide the length  $$b–a$$  into n equal sub intervals, each of length

$$\triangle{x} = \dfrac {b–a}{n}$$ ....(1)

• Now we choose  $$x_1^*,x_2^*,...x_n^*$$  in each sub interval and calculate the average value of the number $$f(x_1^{\ast}), f(x_2^{\ast}),...f(x_n^{\ast})$$

Average = $$\dfrac {f(x_1^{\ast})+f(x_2^{\ast})+....f(x_n^{\ast})}{n}$$

$$=\dfrac {f(x_1^{\ast})+f(x_2^{\ast})+.....f(x_n^{\ast})}{\dfrac{b–a}{\triangle x}}$$

(put  $$n={\dfrac{b–a}{\triangle x}}$$ from (1))

$$\therefore$$  Average  $$=\dfrac {f(x_1^{\ast})\triangle x+f(x_2^{\ast})\triangle x+.....f(x_n^{\ast})\triangle x}{b–a}$$

$$=\dfrac{\sum\limits_{i=1}^nf(x_i^{\ast})\triangle x}{n}$$

• Now, let  $$n$$ increase so that average value of large number of function value is calculated. If we make $$n$$ tends to infinity, we get

Average = $$\lim\limits_{n \to \infty}\dfrac{1}{b–a}\,\sum\limits_{i=1}^nf(x_1^{\ast})\triangle x$$

$$=\dfrac{\int\limits^b_af(x)dx}{b–a}$$

$$\therefore$$ we define the average value of function in the interval $$[a, b]$$ as

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Find the average value of the function,$$f(x) = 2x + x^2$$ in [1, 3].

A $$\dfrac{5}{2}$$

B $$\dfrac{25}{3}$$

C $$\dfrac{17}{4}$$

D $$\dfrac{– 9}{2}$$

×

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Given,

$$f(x) = 2x + x^2 ,\; a = 1, b = 3$$

$$\therefore f_{avg}\dfrac{\int\limits_1^3 (2x+x^2)dx}{3–1}$$

=$$\dfrac{\int\limits_1^3 (2x+x^2)dx}{2}$$

$$\dfrac{\left[2×\dfrac{x^2}{2}+\dfrac{x^3}{3}\right]_1^3}{2}$$

$$\dfrac{\left[x^2+\dfrac{x^3}{3}\right]_1^3}{2}$$

$$\dfrac{(9+9)–(1 + 1/3)}{2}$$$$\dfrac{50}{6}$$

=$$\dfrac{25}{3}$$

Find the average value of the function,$$f(x) = 2x + x^2$$ in [1, 3].

A

$$\dfrac{5}{2}$$

.

B

$$\dfrac{25}{3}$$

C

$$\dfrac{17}{4}$$

D

$$\dfrac{– 9}{2}$$

Option B is Correct

Find the average value of a function $$f(x) = sin^2x\,\, cosx$$  in  $$\big[0,\dfrac \pi2\big]$$

A $$\dfrac{2}{3\pi}$$

B $$\dfrac{3}{2\pi}$$

C $$\dfrac{1}{\pi}$$

D $$\dfrac{–2}{\pi}$$

×

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Given,

$$f(x) = sin^2x\,\, cosx$$ ,  $$b =\dfrac{\pi}{2},\; a = 0$$,

$$f_{average}=\dfrac{\int\limits^{\pi/2}_0sin^2x\;cosx\; dx}{\dfrac{\pi}{2}–0}$$

$$=\dfrac{\int\limits^{\pi/2}_0sin^2x\;cosx\;dx}{\dfrac{\pi}{2}}$$

Now consider

$$I=\int\limits^{\pi/2}_{0}sin^2x\; cosx\; dx$$

put $$sin\,x=t$$

$$\Rightarrow\,\,cos\,x\,dx=dt$$

when $$x = \dfrac{\pi}{2}$$    $$\rightarrow$$  $$t = 1$$

$$x = 0\rightarrow t = 0$$

$$\therefore$$ $$I =\int\limits^{1}_{0} t^2\;dt$$

$$\left[\dfrac{t^3}{3}\right]_0^1$$

$$\dfrac{1}{3}$$

$$\therefore$$  $$f_{average}=\dfrac{1}{3×\dfrac{\pi}{2}}$$

$$=\dfrac{2}{3\pi}$$

Find the average value of a function $$f(x) = sin^2x\,\, cosx$$  in  $$\big[0,\dfrac \pi2\big]$$

A

$$\dfrac{2}{3\pi}$$

.

B

$$\dfrac{3}{2\pi}$$

C

$$\dfrac{1}{\pi}$$

D

$$\dfrac{–2}{\pi}$$

Option A is Correct

Finding the Endpoint of Interval when Average Value of a Function is given in that Interval

• Consider ,

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$ = average value of $$f$$ in [a, b]

If $$f_{(avg)}$$ is known then,

$$(b–a)=f_{(avg)}×\int\limits_a^b\; f(x)\,dx$$

• If we are given one of the two values, 'a' and 'b', we can find the other by using the above  equation.

Find the value of 'b' such that the average value of the function $$f(x)=2x^2-x+1$$ in the internal [0, b] is 2.

A $$b=\dfrac {3+\sqrt {105}}{8}$$

B $$b=\dfrac {2+\sqrt {50}}{3}$$

C $$b=\dfrac {1\pm\sqrt {2}}{4}$$

D $$b=7$$

×

$$f_{(avg)}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Given, $$f_{(avg)}=2$$, b  = ? ,  a = 0

$$f(x)=2x^2-x+1$$

$$\therefore\;$$  $$2×(b-0)=\int\limits_a^b(2x^2-x+1) dx$$

$$\Rightarrow\;2b=\Big [\dfrac {2x^3}{3}-\dfrac {x^2}{2}+x\Big ]_0^b$$

$$\Rightarrow\;2b=\dfrac {2b^3}{3}-\dfrac {b^2}{2}+b$$

$$\Rightarrow\;\dfrac {2b^3}{3}-\dfrac {b^2}{2}-b=0$$

$$\Rightarrow\;\dfrac {2b^2}{3}-\dfrac {b}{2}-1=0$$

$$\Rightarrow\;4b^2-3b-6=0$$

$$\Rightarrow\;b=\dfrac {3\pm\sqrt {9+4×4×6}}{2×4}$$

$$\Rightarrow\;b=\dfrac {3\pm\sqrt {105}}{8}$$  (as $$b > 0$$ , hence $$b=\dfrac {3-\sqrt {105}}{8}$$ is rejected)

$$\Rightarrow\;b=\dfrac {3+\sqrt {105}}{8}$$

Find the value of 'b' such that the average value of the function $$f(x)=2x^2-x+1$$ in the internal [0, b] is 2.

A

$$b=\dfrac {3+\sqrt {105}}{8}$$

.

B

$$b=\dfrac {2+\sqrt {50}}{3}$$

C

$$b=\dfrac {1\pm\sqrt {2}}{4}$$

D

$$b=7$$

Option A is Correct

Mean Value Theorem for Integrals

• If $$f$$ is a continuous function in [a, b] then there exists a number 'c' in [a,b] such that

$$f(c)=f_{(avg)}=\dfrac {1}{b-a}\;\int\limits_a^b\;f(x)dx$$

i.e. $$\int\limits_a^b\;f(x)dx=f(c)(b-a)$$

• This means that average value of a continuous function must be taken at least once by the function in that interval.  Let $$f$$ be a continuous function and $$\int\limits_2^5\;f(x)\,dx=6$$ , which of the following values must be taken by $$f$$ at least once on the interval [2, 5].

A 2

B 5

C –3

D 7

×

Average value must be taken at least once in the interval.

$$f_{(avg)}=\dfrac {\int\limits_a^b\;f(x) \,dx}{b-a}$$

In this case

$$\therefore$$ $$f_{(avg)}=\dfrac {\int\limits_2^5\;f(x) \,dx}{5-2}=\dfrac {6}{3}=2$$

Let $$f$$ be a continuous function and $$\int\limits_2^5\;f(x)\,dx=6$$ , which of the following values must be taken by $$f$$ at least once on the interval [2, 5].

A

2

.

B

5

C

–3

D

7

Option A is Correct

Find the average value of the function  $$f(x) = x\,e^x$$ in the interval [1,3].

A $$3e^3$$

B $$\dfrac{1}{e^3}$$

C $$2e^3$$

D $$e^3$$

×

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Given,

$$f(x) = x\, e^x ,\; a = 1, b = 3$$

$$f_{average}=\dfrac{\int\limits_1^3 x\,e^xdx}{3–1}$$

$$=\dfrac{\int\limits_1^3 x\,e^xdx}{2}$$

Now consider

$$I=\int x\,e^xdx$$ $$\rightarrow$$ use integration by parts

$$I=\int u\,dv = uv\, – \int v\,du.$$

Here, $$u = x,\, dv = e^x\,dx$$

$$\Rightarrow\,v= e^x$$

$$\therefore\;I = x\,e^x –\int e^xdx = x\,e^x–e^x + C$$

$$\therefore$$ $$\int\limits^3_1\,x\,e^xdx$$

$$\left[x\,e^x –e^x\right]_1^3$$

$$= (3e^3–e^3) – (e –e)$$

$$= 2e^3$$

$$f_{average}=\dfrac{2e^3}{2}$$

$$= e^3$$

Find the average value of the function  $$f(x) = x\,e^x$$ in the interval [1,3].

A

$$3e^3$$

.

B

$$\dfrac{1}{e^3}$$

C

$$2e^3$$

D

$$e^3$$

Option D is Correct

Find the average value of the function $$f(x) = \dfrac {sec^2x}{1+tanx}$$ in the interval $$\left [ 0, \dfrac {\pi}{4} \right ]$$.

A $$\dfrac {2\,\ell n\,4}{\pi}$$

B $$\dfrac {4\,\ell n\,2}{\pi}$$

C $$\dfrac {5\,\ell n\,2}{3}$$

D $$\dfrac {5}{3}$$

×

$$f_{average}=\dfrac{\int\limits_a^b f(x)dx}{b–a}$$

Given,  $$f(x) = \dfrac {sec^2x}{1+tanx}$$ , $$a = 0$$, $$b=\dfrac {\pi}{4}$$

$$f_{(avg)}=\dfrac {\int\limits_0^{\pi/4}\; \dfrac {sec^2x}{1+tanx}dx}{\dfrac {\pi}{4}-0}$$

$$=\dfrac {4}{\pi}\int\limits_0^{\pi/4}\; \dfrac {sec^2x}{1+tanx}dx$$

Now consider,

$$I =\int\limits_0^{\pi/4} \dfrac {sec^2x}{1+tan\,x} dx$$

put,  $$1+tan\,x=t$$

$$\Rightarrow sec^2x \,dx = dt$$

when,    $$x=\dfrac {\pi}{4}\rightarrow t=2$$

$$x=0\rightarrow t=1$$

$$\therefore \;I=\int\limits_1^2\dfrac {{dt}} {{t}}= \Big [ \ell n\,t\Big]_1^2$$

$$=\ell n2-\ell n1=\ell n2$$

$$\therefore \; f_{avg}=\dfrac {\ell n2}{\dfrac {\pi}{4}}= \dfrac {4\;\ell n2}{\pi}$$

Find the average value of the function $$f(x) = \dfrac {sec^2x}{1+tanx}$$ in the interval $$\left [ 0, \dfrac {\pi}{4} \right ]$$.

A

$$\dfrac {2\,\ell n\,4}{\pi}$$

.

B

$$\dfrac {4\,\ell n\,2}{\pi}$$

C

$$\dfrac {5\,\ell n\,2}{3}$$

D

$$\dfrac {5}{3}$$

Option B is Correct

If  $$f(x)=\sqrt {x+2}$$ in [–2, 2], find the value of 'c' such that $$f(c)=f_{(avg)}$$.

A $$c=\dfrac {5}{4}$$

B $$c=\dfrac {-2}{9}$$

C $$c=\dfrac {1}{2}$$

D $$c=\dfrac {-1}{3}$$

×

$$f(c)=f_{(avg)}$$

$$\Rightarrow\;\int\limits_a^b\;f(x)dx=f(c)\;(b-a)$$

Given, $$f(x)=\sqrt {x+2},\;\; a =\, –2, b = 2$$,

$$\therefore$$ $$\int\limits_{-2}^{2}\;\sqrt {x+2}\; dx = f(c)=(2-(-2))$$

$$\Rightarrow\; \left [ \dfrac {(x+2)^{3/2}}{3/2} \right]_{-2}^2=f(c)(4)$$

$$\Rightarrow\; \dfrac {2}{3} \left [ 4^{3/2}-0 \right]=4f(c)$$

$$\Rightarrow\; \dfrac {16}{3}=4f(c)$$

$$\Rightarrow\; f(c)=\dfrac {4}{3}$$

$$\therefore\;\sqrt {c+2}=\dfrac {4}{3}$$

$$\Rightarrow\; {c+2}=\dfrac {16}{9}$$

$$\Rightarrow\; c=\dfrac {16}{9}-2$$

$$\Rightarrow\; c=\dfrac {-2}{9}$$

If  $$f(x)=\sqrt {x+2}$$ in [–2, 2], find the value of 'c' such that $$f(c)=f_{(avg)}$$.

A

$$c=\dfrac {5}{4}$$

.

B

$$c=\dfrac {-2}{9}$$

C

$$c=\dfrac {1}{2}$$

D

$$c=\dfrac {-1}{3}$$

Option B is Correct

Finding Average Values of Functions which occur in Practical Life

• Suppose the temperature at a particular place can be expressed as a function of time (i.e. what time of day it is ), then the average value of temperature can be found for that place.
• If  $$T(t)$$  is given then,

$$T_{(avg)}=\dfrac {\int\limits_{t_1}^{t_2}\;T(t)\; dt}{t_2-t_1}$$

where, $$t_1$$ and $$t_2$$ are appropriate beginning and end timings.

At  a certain place the temperature (in °F) t hours after 8 am is given by $$T(t)=60+15\,sin\dfrac {\pi\, t}{12}$$ Find the average temperature during the period 8 am to 8 pm.

A 85°F

B 92°F

C 82.554°F

D 69.554°F

×

$$T_{(avg)}=\dfrac {\int\limits_{t_1}^{t_2}\;T(t)\; dt}{t_2-t_1}$$

where, $$t_1$$ and $$t_2$$ are appropriate beginning and end timings.

In this case,

$$t_1=0$$ (8 am)

$$t_2=12$$ (8 pm)

$$T(t)=60+15\sin\dfrac {\pi\, t}{12}$$

$$\therefore \;T_{(avg)}=\dfrac {\int\limits_{0}^{12}\;\left ( 60+15sin\dfrac {\pi t}{12}\right)\; dt}{12-0}$$

$$=\dfrac {1}{12} \left [60t-\dfrac {15cos\dfrac {\,\pi t}{12}} {{\pi/12}} \right]_0^{12}$$

$$=\dfrac {1}{12} \left [\Big(60×12- {15\,cos\,\pi ×} \dfrac {12}{\pi}\Big)- \Big(0- {15×} \dfrac {12}{\pi}×1\Big) \right]$$

$$=\dfrac {1}{12} \left [720+ \dfrac {180}{\pi}+\dfrac {180}{\pi}\right]$$

$$=\dfrac {1}{12} \left [720+ \dfrac {360}{\pi}\right]$$

$$\Rightarrow\;60+ \dfrac {30}{\pi}$$

$$=60+\dfrac {30}{3.14}=69.554°F$$

At  a certain place the temperature (in °F) t hours after 8 am is given by $$T(t)=60+15\,sin\dfrac {\pi\, t}{12}$$ Find the average temperature during the period 8 am to 8 pm.

A

85°F

.

B

92°F

C

82.554°F

D

69.554°F

Option D is Correct