Average Value of a Function

What You’ll Learn

In this lesson you’ll learn how to find the average value of a function over an interval using a definite integral.

The Concept

You know how to average a list of numbers: add them up and divide by how many there are. But what if the quantity is changing continuously? You can’t just add up finitely many values. The integral handles this.

The average value of f(x) on [a, b] is:

$$f_{\text{avg}} = \frac{1}{b - a} \int_a^b f(x) \, dx$$

The integral computes the total accumulation, and dividing by the interval length (b - a) gives the average height.

Think of it this way: imagine flattening the area under the curve into a rectangle with the same base (b - a). The height of that rectangle is the average value. The rectangle has the same area as the region under the curve, just spread out evenly.

f(x) = x² area under curve = 9 rectangle: 3 x 3 = 9 (same area)

The blue shaded region is the area under x² from 0 to 3 (area = 9). The dashed orange rectangle has base 3 and height 3, so its area is also 9. The height of that rectangle (3) is the average value. The curve’s area got “flattened” into a rectangle of equal area.

Worked Example

Example 1: Average of x² on [0, 3]

$$f_{\text{avg}} = \frac{1}{3 - 0}\int_0^3 x^2 \, dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3$$

The average value of x² on [0, 3] is 3. The function ranges from 0 to 9 on this interval, and the average height is 3.

Example 2: Average velocity

A car’s velocity is v(t) = 3t² + 2t on [0, 4]. Find the average velocity.

$$v_{\text{avg}} = \frac{1}{4}\int_0^4 (3t^2 + 2t) \, dt = \frac{1}{4}\left[t^3 + t^2\right]_0^4 = \frac{1}{4}(64 + 16) = \frac{80}{4} = 20$$

The average velocity is 20 units per time. The car was sometimes going slower and sometimes faster, but 20 is the speed that, if held constant, would cover the same total distance.

Example 3: Average temperature

Temperature over 12 hours is T(t) = 60 + 10t - t² (in degrees, t in hours). Find the average temperature from t = 0 to t = 10.

$$T_{\text{avg}} = \frac{1}{10}\int_0^{10}(60 + 10t - t^2)\,dt = \frac{1}{10}\left[60t + 5t^2 - \frac{t^3}{3}\right]_0^{10}$$ $$= \frac{1}{10}\left(600 + 500 - \frac{1000}{3}\right) = \frac{1}{10}\left(\frac{1800 + 1500 - 1000}{3}\right) = \frac{1}{10} \cdot \frac{2300}{3} \approx 76.7°$$

Real-World Application

Average value shows up whenever you need to summarize a changing quantity:

• Physics: average velocity over a trip, average force during an impact
• Economics: average cost per unit over a production run
• Medicine: average drug concentration in the bloodstream over a dosing period
• Game design: average damage per second (DPS) to balance weapons
• Climate: average temperature over a day, month, or year

You’ve Got This

The average value formula is just “total divided by length.” The integral gives the total, and (b - a) is the length. If you can compute a definite integral, you can find the average value. The geometric picture (flattening the curve into a rectangle of equal area) makes it intuitive.

Quiz

The average value formula is:

• A.f(b) - f(a)
• B.(1/(b-a)) times the integral from a to b of f(x) dx
• C.The integral from a to b of f(x) dx
• D.f'(b) - f'(a)

The average value of x² on [0, 3] is:

• A.9
• B.4.5
• C.3
• D.6

Geometrically, the average value is:

• A.The maximum height of the curve
• B.The height of a rectangle with the same area as the region under the curve
• C.The slope of the secant line
• D.The x-intercept

If the average value of f on [2, 6] is 10, then the integral from 2 to 6 of f(x) dx is:

• A.10
• B.2.5
• C.40
• D.20

Average value is useful in game design for:

• A.Only drawing graphics
• B.Balancing DPS and summarizing changing quantities
• C.Only collision detection
• D.Only scoring