Average Value of a Function: Videos & Practice Problems

The average value of a function f(x) over an interval [a, b] can be calculated using the formula: $\left(\frac{1}{(b-a)}\right)(\int f^{\left(}dx)$. This approach simplifies the process of finding the average output of a function by integrating over the interval and dividing by the interval's length, thus providing a clear method to analyze functions efficiently.

Average Value of a Function

The definite integral is a powerful tool in calculus that allows us to find the average value of a function over a specified interval. To understand this concept, we can start by recalling the process of Riemann sums, which approximate the area under a curve by dividing it into rectangles. As the number of rectangles increases (approaching infinity), the Riemann sum converges to the definite integral, which represents the true area under the curve.

To find the average value of a function \( f(x) \) over the interval \([a, b]\), we can use the formula:

\[\text{Average Value} = \frac{1}{b - a} \int_a^b f(x) \, dx\]

In this formula, \( b - a \) represents the width of the interval, and the integral \( \int_a^b f(x) \, dx \) calculates the total area under the curve from \( a \) to \( b \). The average value thus gives us a single output that represents the function's behavior over the entire interval.

To derive this formula, we start by considering the sum of function outputs at various points within the interval. If we denote the number of subintervals as \( n \), the width of each subinterval can be expressed as \( \Delta x = \frac{b - a}{n} \). The average value can then be approximated by summing the function values at these points and dividing by \( n \). As \( n \) approaches infinity, this sum becomes the definite integral, leading us to the average value formula.

For example, to find the average value of the function \( f(x) = x + 2 \) over the interval \([0, 4]\), we apply the formula:

\[\text{Average Value} = \frac{1}{4 - 0} \int_0^4 (x + 2) \, dx\]

Calculating the integral, we find:

\[\int (x + 2) \, dx = \frac{x^2}{2} + 2x\]

Evaluating this from \( 0 \) to \( 4 \) gives:

\[\left[ \frac{4^2}{2} + 2(4) \right] - \left[ \frac{0^2}{2} + 2(0) \right] = \left[ 8 + 8 \right] - 0 = 16\]

Thus, the average value is:

\[\text{Average Value} = \frac{1}{4} \cdot 16 = 4\]

This example illustrates how to apply the average value formula effectively. Understanding the derivation and application of this formula is crucial for solving problems related to the average value of functions. Practice with various functions and intervals will help solidify this concept.

Find the average value of the function on the interval $[2,5]$.
$F\left(x\right)=x^3-\frac{2}{\sqrt{x}}$

• A. 29.79
• B. 49.65
• C. 74.48
• D. 107.76

Answer: (B)

Find the average value of the function on the interval $[4,9]$.
$F\left(x\right)=\sqrt{x}+6$

• A. 11.26
• B. 4.74
• C. 10.67
• D. 8.53

Answer: (D)

Find the average value of the function on the interval $[0,1]$.
$G\left(x\right)=\frac{2}{x^2+1}$

• A. $\frac{\pi }{4}$
• B. $\pi$
• C. $\frac{\pi }{2}$
• D. $0$

Answer: (C)

Average Value of a Function Example 1

To find the average value of the function \( g(x) = 3x^2 - 2x + 10 \) over the interval from \( a = 2 \) to \( b = 4 \), we use the formula for the average value of a function:

\[\text{Average Value} = \frac{1}{b - a} \int_{a}^{b} g(x) \, dx\]

Substituting the values of \( a \) and \( b \), we have:

\[\text{Average Value} = \frac{1}{4 - 2} \int_{2}^{4} (3x^2 - 2x + 10) \, dx\]

This simplifies to:

\[\text{Average Value} = \frac{1}{2} \int_{2}^{4} (3x^2 - 2x + 10) \, dx\]

Next, we calculate the integral. The antiderivative of \( g(x) \) can be found using the power rule:

\[\int (3x^2 - 2x + 10) \, dx = x^3 - x^2 + 10x + C\]

Now, we evaluate this from \( 2 \) to \( 4 \):

\[\left[ x^3 - x^2 + 10x \right]_{2}^{4}\]

Calculating the upper bound at \( x = 4 \):

\[4^3 - 4^2 + 10 \cdot 4 = 64 - 16 + 40 = 88\]

Now, calculating the lower bound at \( x = 2 \):

\[2^3 - 2^2 + 10 \cdot 2 = 8 - 4 + 20 = 24\]

Subtracting the lower bound from the upper bound gives:

\[88 - 24 = 64\]

Now, substituting back into the average value formula:

\[\text{Average Value} = \frac{1}{2} \cdot 64 = 32\]

Thus, the average value of the function \( g(x) \) on the interval from \( 2 \) to \( 4 \) is \( 32 \).

To visualize this, if we were to sketch the graph of \( g(x) = 3x^2 - 2x + 10 \), it would be a parabola opening upwards. The vertex would be above the x-axis, and the function would increase as \( x \) moves away from the vertex. The average value of \( 32 \) can be indicated on the graph as a horizontal line intersecting the y-axis at \( 32 \) between the points corresponding to \( x = 2 \) and \( x = 4 \). This line represents the average value of the function over the specified interval.

Here’s what students ask on this topic:

What is the formula for finding the average value of a function over an interval?

The formula for finding the average value of a function f(x) over an interval [a, b] is given by:

$A=\frac{1}{(b-a)}\int a_b^{}f\left(x\right)dx$

This formula uses the definite integral to calculate the average output of the function across the interval [a, b]. It simplifies the process by leveraging Riemann sums and the concept of limits as the number of subintervals approaches infinity.

How do you derive the equation for the average value of a function using Riemann sums?

To derive the equation for the average value of a function using Riemann sums, start by considering the sum of function outputs over subintervals. The average value is the sum of outputs divided by the number of subintervals. As the number of subintervals approaches infinity, this sum becomes a definite integral. The width of each subinterval is Δx = (b - a)/n. Rearranging gives n = (b - a)/Δx. Substitute this into the average value formula, and as n approaches infinity, the sum becomes the integral from a to b. The final formula is:

$A=\frac{1}{(b-a)}\int a_b^{}f\left(x\right)dx$

How can you find the average value of the function f(x) = x + 2 on the interval [0, 4]?

To find the average value of the function f(x) = x + 2 on the interval [0, 4], use the formula:

$A=\frac{1}{(4-0)}\int 0_4^{}(x+2)dx$

Calculate the integral of x + 2 from 0 to 4. The antiderivative of x is x²/2, and the antiderivative of 2 is 2x. Evaluate these from 0 to 4:

$\frac{1}{4}[\frac{4}{2}+8-0]$

This simplifies to 4, which is the average value of the function on the interval.

Why is the concept of limits important in finding the average value of a function?

The concept of limits is crucial in finding the average value of a function because it allows us to handle the infinite number of possible x-values within an interval. By taking the limit as the number of subintervals approaches infinity, we can accurately calculate the area under the curve using Riemann sums, which then translates into the definite integral. This process ensures that we obtain the true average value of the function over the specified interval, as it accounts for all possible inputs and outputs, providing a precise and comprehensive result.