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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