Q. 11 - Calculus

Question

The average value of a function f on [a, b] .

Step-by-Step Solution

Answer

$\text{ The average value of function }f\left(x\right)\text{ on interval }[a,b]\text{ is defined as }\overline{f\left(x\right)}=\frac{1}{b-a}{\int }_a^b f\left(x\right)dx\text{. }$

Step 1. Given Information

$\text{ The functions }f\left(x\right)\text{ continuous on interval }[a,b]\text{. }$

Step 2. Formula used

$$\begin{gathered} \text{ The average }\left(\text{ mean }\right)\text{ of function }f\left(x\right)\text{ for arguments }{x}_1,x_2,x_3\dots x_n\text{ is given by }\overline{f\left(x\right)}=\frac{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\dots f\left(x_n\right)}{n}.\text{ The } \\ \text{ integral; }A={\int }_a^b f\left(x\right)dx\text{, estimates the area between the graph of the function }f\left(x\right)\text{ and }x\text{-axis on an interval } \\ [a,b]. \end{gathered}$$

Step 3. Definition/Explanation

$$\begin{gathered} \text{ The graph of function }f\left(x\right)\text{ is continuous on interval }[a,b]\text{. The average value of the function }f\left(x\right)\text{ is } \\ \overline{f\left(x\right)}=\frac{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\dots f\left(x_n\right)}{n}=\frac{\sum _{i=1}^n f\left(x_1\right)}{n}.................\left(1\right) \\ \text{ Now, we divide length of interval }b-a\text{ into n parts such that }\frac{b-a}{n}=\mathrm{\Delta }x,\mathrm{\Delta }x\text{ is as small as we can take it. } \\ \text{ Thercforc, substituting valuc of }n\text{ rcsult (1), wc have.} \\ \overline{f\left(x\right)}=\frac{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)\dots f\left(x_n\right)}{n}=\frac{\sum _{i=1}^n f\left(x_k\right)}{\frac{b-a}{\mathrm{\Delta }x}}=\frac{\sum _{k=1}^n f\left(x_k\right)\mathrm{\Delta }x}{b-a},.\text{ But }f\left(x_k\right)\mathrm{\Delta }x\text{ is the area of small rectangle made by } \\ \text{ vertical length }f\left(x_k\right)\text{ and width }\mathrm{△}x\text{, and so }\sum _{k=1}^n f\left(x_k\right)\mathrm{\Delta }x\text{ is the sum of areas of all rectangles on the interval } \\ [a,b]\text{. That further implies }\sum _{k=1}^n f\left(x_k\right)\mathrm{\Delta }x\text{ is the area between the graph of function }f\left(x\right)\text{ and }x\text{-axis on an interval [a,b].} \\ \text{ Since, }f\left(x\right)\text{ is a continuous function, therefore, we have }\sum _{k=1}^n f\left(x_k\right)\mathrm{\Delta }x=\int f\left(x\right)dx\text{. } \\ \Rightarrow \overline{f\left(x\right)}=\frac{\sum _{i=1}^n f\left(x_k\right)\mathrm{\Delta }x}{b-a}=\frac{1}{b-a}{\int }_a^b f\left(x\right)dx \\ \text{ Thus, the average value of function is }f(x{)}_{av}=\overline{f\left(x\right)}=\frac{1}{b-a}{\int }_a^b f(x)dx\text{. } \end{gathered}$$

Step 4. Conclusion

$\text{ The average value of function }f\left(x\right)\text{ on interval }[a,b]\text{ is defined as }\overline{f\left(x\right)}=\frac{1}{b-a}{\int }_a^b f\left(x\right)dx\text{. }$