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

- 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)

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

- 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

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

A \(\dfrac{5}{2}\)

B \(\dfrac{25}{3}\)

C \(\dfrac{17}{4}\)

D \(\dfrac{– 9}{2}\)

A \(\dfrac{2}{3\pi}\)

B \(\dfrac{3}{2\pi}\)

C \(\dfrac{1}{\pi}\)

D \(\dfrac{–2}{\pi}\)

- 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.

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

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

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

D \(b=7\)

- 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.

A \(3e^3\)

B \(\dfrac{1}{e^3}\)

C \(2e^3\)

D \(e^3\)

A \(\dfrac {2\,\ell n\,4}{\pi}\)

B \(\dfrac {4\,\ell n\,2}{\pi}\)

C \(\dfrac {5\,\ell n\,2}{3}\)

D \(\dfrac {5}{3}\)

A \(c=\dfrac {5}{4}\)

B \(c=\dfrac {-2}{9}\)

C \(c=\dfrac {1}{2}\)

D \(c=\dfrac {-1}{3}\)

- 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.