Question

In Problems 3 to 12, find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the x axis are the same.

$\mathbf{1}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{o}\mathit{n}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{)}$

Short answer

The average value of the function $1-e^{-x}$on (0, 1) is calculated to be $e^{-1}$. It means that the area on the upper half of the curve of the function is greater than the lower half of the curve of the function.

Step-by-step solution

Definition of amplitude, period, frequency, and velocity amplitude.

The average value of the function in the interval (a, b)is defined as.

$\mathit{f}\mathbf{(}\mathbf{x}{\mathbf{)}}_{\mathbf{A}\mathbf{v}\mathbf{g}}\mathbf{=}\frac{{\mathbf{\int }}_{\mathbf{a}}^{\mathbf{b}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{.}\mathbf{d}\mathbf{x}}{\mathbf{b}\mathbf{-}\mathbf{a}}$

Integration of sine function is$\mathbf{\int }\mathit{s}\mathit{i}\mathit{n}\left(ax+b\right)\mathbf{.}\mathit{d}\mathit{x}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{a}}\mathit{c}\mathit{o}\mathit{s}\left(ax+b\right)$.

Given parameters

The given function is $\text{1}-{\text{e}}^{-x}$.

The average value of function on interval (1, 0) is to be found.

Calculation of average value of function in given interval

Integrate given equation with upper limit be 1 and lower limit be 0. Use the formulae to calculate the average value of function f(x).

$$\begin{gathered} f(x{)}_{Avg}=\underset{0}{\overset{1}{\int }}\left(1-e^{-x}\right).dx \\ f(x{)}_{Avg}={\left[x\right]}_0^1-{\left[\frac{e^{-x}}{-1}\right]}_0^1 \\ f(x{)}_{Avg}=1-0+\left(e^{-1}-e^0\right) \\ f(x{)}_{Avg}=e^{-1} \end{gathered}$$

Hence, the average value of the function $\text{1}-e^{-x}$on (0, 1) is $e^{-1}$.

Graph of the function

The graph of the function $\text{1}-{\text{e}}^{-x}$in the interval (0, 1) using graphing calculator

Therefore, it is clear from the graph that the area subtends by the function above the X-axis is greater than the area subtended below the X-axis.