6.7 Average Value Of A Functionap Calculus

  1. 6.7 Average Value Of A Functional Calculus Calculator
  2. 6.7 Average Value Of A Functional Calculus Equation
  3. 6.7 Average Value Of A Functional Calculus 2nd Edition
  4. 6.7 Average Value Of A Functional Calculus 14th Edition

Definition of Average Value

6.7 The formula for average value of a continuous function F(x) on a, b is 1/(b - a) inta^b F(x)dx Our formula here will be A = 1/(4 - 1) int1^4 2^x dx A = 1/3 int1^4 2^x dx A = 1/32^x/ln21^4 A = 1/32^4/ln2 - 2^1/ln2 A = 1/314/ln2 A = 14/(3ln2) This can be approximated as 6.7. The average value of y = 2^x on 1, 4 is 6.7. Hopefully this helps! Assignments AB 2020-2021. 1.09 Average Vs. Instantaneous Rates. 1.10 Secant, Tangent, and Normal Lines. Mean Value Theorem.

One of the main applications of definite integrals is to find the average value of a function (y = fleft( x right)) over a specific interval (left[ {a,b} right].)

In order to find this average value, one must integrate the function by using the Fundamental Theorem of Calculus and divide the answer by the length of the interval.

So, the average (or the mean) value of (fleft( x right)) on (left[ {a,b} right]) is defined by

[bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} .]

The Mean Value Theorem for Definite Integrals

Let (y = fleft( x right)) be a continuous function on the closed interval (left[ {a,b} right].) The mean value theorem for integrals states that there exists a point (c) in that interval such that

[fleft( c right) = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} .]

In other words, the mean value theorem for integrals states that there is at least one point (c) in the interval (left[ {a,b} right]) where (fleft( x right)) attains its average value (bar f:)

[{fleft( c right) = bar f }={ frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} .}]

Geometrically, this means that there is a rectangle whose area exactly represents the area of the region under the curve (y = fleft( x right).) The value of (fleft( c right)) represents the height of the rectangle and the difference (left( {b – a} right)) represents the width.

Root Mean Square Value of a Function

The root mean square value (left( {RMS} right)) is defined as the square root of the average (mean) value of the squared function ({{{left[ {fleft( x right)} right]}^2}}) over an interval (left[ {a,b} right].) The corresponding integration formula is written in the form

[RMS = sqrt {frac{1}{{b – a}}intlimits_a^b {{{left[ {fleft( x right)} right]}^2}dx} } .]

The ({RMS}) value has many applications in mathematics, physics and engineering. For example, in physics, the RMS value of an alternating current (left( {AC} right)) is equal to the value of the direct current (left( {DC} right)) that dissipates the same power in a resistor.

Solved Problems

Click or tap a problem to see the solution.

Example 1

The average value of a function (y = fleft( x right)) over the interval (x in left[ {1,5} right]) is (2.) What is the value of (intlimits_1^5 {fleft( x right)dx}?)

Example 2

Find the average value of the cubic function (fleft( x right) = {x^3}) on the interval (left[ {0, 1} right].)

Example 3

Find the average value of the square root function (fleft( x right) = sqrt{x}) on the interval (left[ {0, 25} right].)

Example 4

Find the average value of the cosine function (fleft( x right) = cos{x}) on the interval (left[ {0, large{frac{pi }{2}}normalsize} right].)

Example 5

Find the average value of the sine squared function (fleft( x right) = {sin ^2}x) on the interval (left[ {0,pi } right].)
6.7 Average Value Of A Functionap Calculus

Example 6

The daily temperature of the outside air is given by the equation[Tleft( t right) = 20 – 5cos left( {frac{{pi t}}{{12}}} right),]where (t) is measured in hours (left( {0 le t le 24} right)) and (T) is measured in degrees (C.) Find the average temperature between ({t_1} = 6) and ({t_2} = 12) hours.

Example 7

Given the rational function[fleft( x right) = frac{2}{{{{left( {x + 1} right)}^2}}}.]Find the values of (c) that satisfy the Mean Value Theorem for Integrals for the function on the interval (left[ {0,3} right].)

Example 8

Given the quadratic function[fleft( x right) = {left( {x + 2} right)^2}.]Find the values of (c) that satisfy the Mean Value Theorem for Integrals for the function on the interval (left[ {0,9} right].)

Example 9

A sawtooth signal has the period (T = 1) and is given by the equation (fleft( t right) = Atleft({text{mod} A}right).) Find the RMS value of the sawtooth waveform.

Example 10

Find the RMS value of the sine function (fleft( t right) = Asin t) over the interval (left[ {0,2pi } right].)

Example 1.

The average value of a function (y = fleft( x right)) over the interval (x in left[ {1,5} right]) is (2.) What is the value of (intlimits_1^5 {fleft( x right)dx}?)

Solution.

By definition,

6.7 Average Value Of A Functional Calculus Calculator

[{bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} ,};; Rightarrow {intlimits_a^b {fleft( x right)dx} = bar fleft( {b – a} right).}]

Substituting the given values, we obtain

[{intlimits_1^5 {fleft( x right)dx} }={ 2left( {5 – 1} right) }={ 8.}]

Example 2.

Find the average value of the cubic function (fleft( x right) = {x^3}) on the interval (left[ {0, 1} right].)

Solution.

We use the integration formula

[bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} .]

Hence,

[{bar f = frac{1}{{1 – 0}}intlimits_0^1 {{x^3}dx} }={ intlimits_0^1 {{x^3}dx} }={ left. {frac{{{x^4}}}{4}} right _0^1 }={ frac{1}{4}.}]

Example 3.

Find the average value of the square root function (fleft( x right) = sqrt{x}) on the interval (left[ {0, 25} right].)

Solution.

Using the definition of the average value, we can write

[{bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} }={ frac{1}{{25}}intlimits_0^{25} {sqrt x dx} }={ frac{1}{{25}}intlimits_0^{25} {{x^{frac{1}{2}}}dx} }={ frac{1}{{25}} cdot left. {frac{{2{x^{frac{3}{2}}}}}{3}} right _0^{25} }={ frac{1}{{25}} cdot frac{{2{{left( {sqrt {25} } right)}^3}}}{3} }={ frac{{250}}{{75}} }={ frac{{10}}{3}.}]

Example 4.

Find the average value of the cosine function (fleft( x right) = cos{x}) on the interval (left[ {0, large{frac{pi }{2}}normalsize} right].)

Solution.

6.7 Average Value Of A Functional Calculus Equation

We use the formula

[bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} .]

So we set up and evaluate the following integral

[{bar f = frac{1}{{frac{pi }{2} – 0}}intlimits_0^{frac{pi }{2}} {cos xdx} }={ frac{2}{pi }intlimits_0^{frac{pi }{2}} {cos xdx} }={ frac{2}{pi }left. {sin x} right _0^{frac{pi }{2}} }={ frac{2}{pi }.}]

Example 5.

Find the average value of the sine squared function (fleft( x right) = {sin ^2}x) on the interval (left[ {0,pi } right].)

Solution.

The average value (bar f) is given by

[bar f = frac{1}{pi }intlimits_0^pi {{{sin }^2}xdx} .]

Using the trigonometric identity

[{sin ^2}x = frac{1}{2}left( {1 – cos 2x} right),]

we obtain

[require{cancel}{bar f = frac{1}{{2pi }}intlimits_0^pi {left( {1 – cos 2x} right)dx} }={ frac{1}{{2pi }}left. {left[ {x – frac{{sin 2x}}{2}} right]} right _0^pi }={ frac{1}{{2pi }}left[ {left( {pi – 0} right) – left( {0 – 0} right)} right] }={ frac{cancel{pi} }{{2 cancel{pi} }} }={ frac{1}{2}.}]

Example 6.

The daily temperature of the outside air is given by the equation[Tleft( t right) = 20 – 5cos left( {frac{{pi t}}{{12}}} right),]where (t) is measured in hours (left( {0 le t le 24} right)) and (T) is measured in degrees (C.) Find the average temperature between ({t_1} = 6) and ({t_2} = 12) hours.

Solution.

We calculate the average temperature in the given interval through integration using the definition of the average value of a function:

[{{{bar T}_{left[ {6,12} right]}} = frac{1}{{12 – 6}}intlimits_6^{12} {Tleft( t right)dt} }={ frac{1}{6}intlimits_6^{12} {left[ {20 – 5cos left( {frac{{pi t}}{{12}}} right)} right]dt} }={ frac{1}{6}left. {left[ {20t – 5 cdot frac{{12}}{pi }sin left( {frac{{pi t}}{{12}}} right)} right]} right _6^{12} }={ frac{1}{6}left[ {left( {240 – frac{{60}}{pi }sin pi } right) }right.}-{left.{ left( {120 – frac{{60}}{pi }sin frac{pi }{2}} right)} right] }={ frac{1}{6}left( {120 + frac{{60}}{pi }} right) }={ 20 + frac{{10}}{pi } approx 23.2^text{ o}C}]

Example 7.

Given the rational function[fleft( x right) = frac{2}{{{{left( {x + 1} right)}^2}}}.]Find the values of (c) that satisfy the Mean Value Theorem for Integrals for the function on the interval (left[ {0,3} right].)

Solution.

First we calculate the average value of the function (fleft( x right)) on the interval (left[ {0,3} right]:)

[{bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} }={ frac{1}{3}intlimits_0^3 {frac{{2dx}}{{{{left( {x + 1} right)}^2}}}} }={ frac{2}{3}left. {left( { – frac{1}{{x + 1}}} right)} right _0^3 }={ frac{2}{3}left( {1 – frac{1}{4}} right) }={ frac{1}{2}.}]

To determine the values of (c,) we solve the equation

[ fleft( c right) = bar f .]

Hence,

[{frac{2}{{{{left( {c + 1} right)}^2}}} = frac{1}{2},};; Rightarrow {{left( {c + 1} right)^2} = 4,};; Rightarrow {c + 1 = pm 2,};; Rightarrow {{c_1} = 1,;}kern0pt{{c_2} = – 1.}]

We see that only the positive root ({c_1} = 1) lies in the interval (left[ {0,3} right]), so the answer is (c = 1.)

Example 8.

Given the quadratic function[fleft( x right) = {left( {x + 2} right)^2}.]Find the values of (c) that satisfy the Mean Value Theorem for Integrals for the function on the interval (left[ {0,9} right].)

Solution.

The average value of the function (fleft( x right)) on the interval (left[ {0,9} right]) is given by

[{bar f = frac{1}{{b – a}}intlimits_a^b {fleft( x right)dx} }={ frac{1}{9}intlimits_0^9 {{{left( {x + 2} right)}^2}dx} }={ frac{1}{9}intlimits_0^9 {left( {{x^2} + 4x + 4} right)dx} }={ frac{1}{9}left. {left[ {frac{{{x^3}}}{3} + 2{x^2} + 4x} right]} right _0^9 }={ frac{{441}}{9} }={ 49.}]

To find the values of (c,) we solve the equation (bar f = fleft( c right).) This yields

[{{left( {c + 2} right)^2} = 49,};; Rightarrow {c + 2 = pm 7,};; Rightarrow {{c_1} = – 9,;}kern0pt{{c_2} = 5.}]

We see that the solution is (c = 5.)

6.7 average value of a functional calculus 2nd edition

Example 9.

A sawtooth signal has the period (T = 1) and is given by the equation (fleft( t right) = Atleft({text{mod} A}right).) Find the RMS value of the sawtooth waveform.

Solution.

The sawtooth signal is periodic, so we integrate over one cycle from (t = 0) to (t = 1.) The (RMS) value is expressed by the formula

[{RMS = sqrt {frac{1}{{1 – 0}}intlimits_0^1 {{{left( {At} right)}^2}dt} } }={ sqrt {{A^2}intlimits_0^1 {{t^2}dt} } }={ Asqrt {left. {frac{{{t^3}}}{3}} right _0^1} }={ frac{A}{{sqrt 3 }}.}]

Example 10.

Find the RMS value of the sine function (fleft( t right) = Asin t) over the interval (left[ {0,2pi } right].)

Solution.

The (RMS) value is given by the integration formula

[RMS = sqrt {frac{1}{{2pi }}intlimits_0^{2pi } {{{sin }^2}tdt} } .]

Using the trig identity

[{sin ^2}t = frac{1}{2}left( {1 – cos 2t} right),]

6.7 Average Value Of A Functional Calculus 2nd Edition

we have

6.7 Average Value Of A Functional Calculus 14th Edition

[{RMS = sqrt {frac{1}{{2pi }}intlimits_0^{2pi } {{A^2}{{sin }^2}tdt} } }={ sqrt {frac{1}{{2pi }}intlimits_0^{2pi } {frac{{{A^2}}}{2}left( {1 – cos 2t} right)dt} } }={ sqrt {frac{{{A^2}}}{{4pi }}intlimits_0^{2pi } {left( {1 – cos 2t} right)dt} } }={ sqrt {frac{{{A^2}}}{{4pi }}left. {left[ {t – frac{{sin 2t}}{2}} right]} right _0^{2pi }} }={ sqrt {frac{{{A^2}}}{{4pi }} cdot 2pi } }={ frac{A}{{sqrt 2 }}.}]

The calculator will find the average value of the function on the given interval, with steps shown.

  • In general, you can skip the multiplication sign, so `5x` is equivalent to `5*x`.
  • In general, you can skip parentheses, but be very careful: e^3x is `e^3x`, and e^(3x) is `e^(3x)`.
  • Also, be careful when you write fractions: 1/x^2 ln(x) is `1/x^2 ln(x)`, and 1/(x^2 ln(x)) is `1/(x^2 ln(x))`.
  • If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. write sin x (or even better sin(x)) instead of sinx.
  • Sometimes I see expressions like tan^2xsec^3x: this will be parsed as `tan^(2*3)(x sec(x))`. To get `tan^2(x)sec^3(x)`, use parentheses: tan^2(x)sec^3(x).
  • Similarly, tanxsec^3x will be parsed as `tan(xsec^3(x))`. To get `tan(x)sec^3(x)`, use parentheses: tan(x)sec^3(x).
  • From the table below, you can notice that sech is not supported, but you can still enter it using the identity `sech(x)=1/cosh(x)`.
  • If you get an error, double-check your expression, add parentheses and multiplication signs where needed, and consult the table below.
  • All suggestions and improvements are welcome. Please leave them in comments.
The following table contains the supported operations and functions:
TypeGet
Constants
ee
pi`pi`
ii (imaginary unit)
Operations
a+ba+b
a-ba-b
a*b`a*b`
a^b, a**b`a^b`
sqrt(x), x^(1/2)`sqrt(x)`
cbrt(x), x^(1/3)`root(3)(x)`
root(x,n), x^(1/n)`root(n)(x)`
x^(a/b)`x^(a/b)`
x^a^b`x^(a^b)`
abs(x)` x `
Functions
e^x`e^x`
ln(x), log(x)ln(x)
ln(x)/ln(a)`log_a(x)`
Trigonometric Functions
sin(x)sin(x)
cos(x)cos(x)
tan(x)tan(x), tg(x)
cot(x)cot(x), ctg(x)
sec(x)sec(x)
csc(x)csc(x), cosec(x)
Inverse Trigonometric Functions
asin(x), arcsin(x), sin^-1(x)asin(x)
acos(x), arccos(x), cos^-1(x)acos(x)
atan(x), arctan(x), tan^-1(x)atan(x)
acot(x), arccot(x), cot^-1(x)acot(x)
asec(x), arcsec(x), sec^-1(x)asec(x)
acsc(x), arccsc(x), csc^-1(x)acsc(x)
Hyperbolic Functions
sinh(x)sinh(x)
cosh(x)cosh(x)
tanh(x)tanh(x)
coth(x)coth(x)
1/cosh(x)sech(x)
1/sinh(x)csch(x)
Inverse Hyperbolic Functions
asinh(x), arcsinh(x), sinh^-1(x)asinh(x)
acosh(x), arccosh(x), cosh^-1(x)acosh(x)
atanh(x), arctanh(x), tanh^-1(x)atanh(x)
acoth(x), arccoth(x), cot^-1(x)acoth(x)
acosh(1/x)asech(x)
asinh(1/x)acsch(x)