Average Value of a Function

An economist finds that the gas prices from 2017 to 2022 can be described by the function

\[f(x) = 1.4^x.\]

Here, \( f \) is measured in dollars per gallon, and \(x\) represents the number of years since 2017. Find the average price of gas per gallon between 2017 and 2022.

Answer:

In order to use the formula for the average value of a function you first need to identify the interval. Since the function measures the years since 2017, then the interval becomes \( [0,5],\) where 0 represents 2017 and 5 represents 2022.

Next, you will need to find the definite integral

\[\int_0^5 1.4^x\,\mathrm{d}x.\]

Begin by finding its antiderivative:

\[ \int 1.4^x\,\mathrm{d}x= \frac{1}{\ln{1.4}} 1.4^x,\]

and then use the Fundamental Theorem of Calculus to evaluate the definite integral, giving you

\[ \begin{align} \int_0^5 1.4^x\,\mathrm{d}x &=\left( \frac{1}{\ln{1.4}} 1.4^5 \right) - \left( \frac{1}{\ln{1.4}} 1.4^0 \right) \\ &= \frac{1.4^5-1}{\ln{1.4}} \\ &= 13.012188. \end{align} \]

Now that you found the value of the definite integral, you divide by the length of the interval, so

\[ \begin{align} f_{\text{avg}} &= \frac{13.012188}{5} \\ &= 2.6024376. \end{align}\]

This means that the average price of gas between 2017 and 2022 is $2.60 per gallon.

Take a look at a graphical representation of the problem:

Graphical representation of the average value of the price of the gas

The rectangle represents the total area under the curve of \(f(x)\). The rectangle has a width of \(5\), which is the interval of integration, and a height equal to the average value of the function, \(2.6\).

Find the average value of

\[ g(x) = x^3 \]

in the interval \( [-2,1].\)

Answer:

This time the interval is given in a straightforward way, so begin by finding the indefinite integral

\[ \int x^3 \, \mathrm{d}x, \]

which you can do by using the Power Rule, to find that

\[ \int x^3 \, \mathrm{d}x = \frac{1}{4}x^4.\]

Next, use the Fundamental Theorem of Calculus to evaluate the definite integral. This gives you

\[ \begin{align} \int_{-2}^1 x^3 \, \mathrm{d}x &= \left( \frac{1}{4}(1)^4 \right) - \left( \frac{1}{4} (-2)^4 \right) \\ &= \frac{1}{4} - 4 \\ &= -\frac{15}{4}. \end{align} \]

Finally, divide the value of the definite integral by the length of the interval, so

\[ \begin{align} g_{\text{avg}} &= \frac{1}{1-(-2)}\left(-\frac{15}{4} \right) \\ &= -\frac{15}{12} \\ &= - \frac{5}{4}. \end{align}\]

Therefore, the average value of \( g(x) \) in the interval \( [-2,1] \) is \( -\frac{5}{4}.\)

Find the average value of \(h(x) = x \) on the interval \( [-3,3].\)

Answer:

Begin by using the Power Rule to find the indefinite integral, that is

\[ \int x \, \mathrm{d}x = \frac{1}{2}x^2.\]

Knowing this, you can evaluate the definite integral, so

\[ \begin{align} \int_{-3}^3 x\, \mathrm{d}x &= \left( \frac{1}{2}(3)^2\right)-\left(\frac{1}{2}(-3)^2\right) \\ &= \frac{9}{2}-\frac{9}{2} \\ &= 0. \end{align}\]

Since the definite integral is equal to 0, you will also get 0 after dividing by the length of the interval, so

\[ h_{\text{avg}}=0.\]

Find the average value of

\[f(x) = \sin(x)\]

over the interval \( \left[ 0, \frac{\pi}{2} \right].\)

Answer:

You will need to find first the definite integral

\[ \int_0^{\frac{\pi}{2}} \sin{x} \, \mathrm{d}x,\]

so find its antiderivative

\[ \int \sin{x} \, \mathrm{d}x = -\cos{x},\]

and use the Fundamental Theorem of Calculus to evaluate the definite integral, that is

\[ \begin{align} \int_0^{\frac{\pi}{2}} \sin{x} \, \mathrm{d}x &= \left(-\cos{\frac{\pi}{2}} \right) - \left(-\cos{0} \right) \\ &= -0-\left( -1 \right) \\ &= 1. \end{align}\]

Finally, divide by the length of the interval, so

\[ \begin{align} f_{\text{avg}} &= \frac{1}{\frac{\pi}{2}}\\ &= \frac{2}{\pi}. \end{align}\]

This means that the average value of the sine function over the interval \( \left[ 0, \frac{\pi}{2} \right]\) is \(\frac{2}{\pi},\) which is about \(0.63.\)

Graphical representation of the average value of the sine function in the interval \( [0,\frac{\pi}{2}].\)