8.3 Average Value of a Function & The Mean Value Theorem for Integrals

8.3 Average Value of a Function & The Mean Value Theorem for Integrals

Core CLEP Concept: While finding the average of a finite collection of numbers requires dividing their sum by $n$, computing the average value of a continuously varying function $f(x)$ over an interval $[a, b]$ requires definite integration. The Mean Value Theorem for Definite Integrals guarantees that every continuous function achieves its exact average value at least once on the interval. On the CLEP Calculus exam, you must calculate $f_{\text{avg}}$, find the specific coordinate $c$ where $f(c) = f_{\text{avg}}$, and interpret the theorem geometrically and physically.

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1. Discrete Average vs. Continuous Average

The Arithmetic Mean of Discrete Numbers

For a finite set of $n$ numbers $\{y_1, y_2, \dots, y_n\}$, the standard average (arithmetic mean) is:

$$\bar{y} = \frac{y_1 + y_2 + \dots + y_n}{n} = \frac{1}{n} \sum_{i=1}^n y_i$$

Transition to a Continuous Function

To find the average value of a continuous function $f(x)$ on $[a, b]$, sample the function at $n$ equally spaced points $x_1^*, x_2^*, \dots, x_n^*$ with subinterval width $\Delta x = \frac{b - a}{n}$.

Solving for $n$ gives $n = \frac{b - a}{\Delta x}$. Substituting this into the discrete average formula yields:

$$\text{Average} \approx \frac{1}{n} \sum_{i=1}^n f(x_i^*) = \frac{1}{\frac{b - a}{\Delta x}} \sum_{i=1}^n f(x_i^*) = \frac{1}{b - a} \sum_{i=1}^n f(x_i^*) \, \Delta x$$

Taking the limit as $n \to \infty$ (so $\Delta x \to 0$), the Riemann sum becomes the definite integral:

$$f_{\text{avg}} = \lim_{n \to \infty} \frac{1}{b - a} \sum_{i=1}^n f(x_i^*) \, \Delta x = \frac{1}{b - a} \int_a^b f(x) \, dx$$

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2. Definition & Geometric Interpretation

Definition: Average Value of a Continuous Function

If $f$ is integrable on the closed interval $[a, b]$, then the average value of $f$ on $[a, b]$ is:

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

The Equal-Area Rectangle Principle

Multiplying both sides of the definition by $(b - a)$ reveals its profound geometric meaning:

$$\text{Area of Rectangle} = f_{\text{avg}} \cdot (b - a) = \int_a^b f(x) \, dx = \text{Area Under Curve}$$

   y ^                                                
     |               .---.  y = f(x)                  
     |             .'     '.                          
     |   +--------/---------\--------+  y = f_avg     
     |   | Area  /           \  Area |                
     |   | Void /  Area Equal \ Void |                
     |   |     /   to Integral \     |                
     |   |    /                 \    |                
     |   +---.-------------------.---+                
   --+-------+-------------------+-----------> x      
     0       a         c         b                    
             |<----- (b - a) --->|                    

• The horizontal line $y = f_{\text{avg}}$ defines a rectangle with base width $(b - a)$ and height $f_{\text{avg}}$.
• The area of this rectangle is identical to the net area under the curve $y = f(x)$ from $x = a$ to $x = b$.
• Conceptually, $f_{\text{avg}}$ "levels out" the peaks and valleys: the area of the curve rising above $y = f_{\text{avg}}$ exactly fills the vacant spaces below $y = f_{\text{avg}}$.

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3. The Mean Value Theorem for Definite Integrals

Formal Theorem Statement

If $f$ is continuous on the closed interval $[a, b]$, then there exists at least one number $c \in [a, b]$ such that:

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

Rigorous Mathematical Connection to Differential MVT

The Mean Value Theorem for Integrals is the integral counterpart of the differential MVT. Consider the accumulation function:

$$F(x) = \int_a^x f(t) \, dt$$

1. By the Fundamental Theorem of Calculus Part 1, $F(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$, with $F'(x) = f(x)$.
2. By the Mean Value Theorem for Derivatives, there exists a number $c \in (a, b)$ such that: $$F'(c) = \frac{F(b) - F(a)}{b - a}$$
3. Substituting $F'(c) = f(c)$, $F(b) = \int_a^b f(t) \, dt$, and $F(a) = \int_a^a f(t) \, dt = 0$ yields: $$f(c) = \frac{\int_a^b f(t) \, dt - 0}{b - a} = \frac{1}{b - a} \int_a^b f(x) \, dx$$

Geometric Interpretation of $c$

The graph of $y = f(x)$ must physically intersect the horizontal average value line $y = f_{\text{avg}}$ at the coordinate point $(c, f(c))$ at least once inside the interval $[a, b]$.

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4. Step-by-Step Protocol for Finding $f_{\text{avg}}$ and Solving for $c$

1. Compute the Definite Integral: Evaluate $I = \int_a^b f(x) \, dx$ using antiderivatives.
2. Calculate the Average Value: Divide the integral by the interval length: $f_{\text{avg}} = \frac{I}{b - a}$.
3. Set Up the Algebraic Equation: Equate the function to its average: $f(c) = f_{\text{avg}}$.
4. Solve for $c$: Find all algebraic solutions for $c$.
5. Validate Boundaries: Discard any roots of $c$ that do not lie within the closed interval $[a, b]$.

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5. Step-by-Step Worked Examples

Worked Example 1: Quadratic Polynomial

Problem: For the function $f(x) = 3x^2 - 2x + 1$ on the interval $[0, 3]$:

• (a) Compute the average value $f_{\text{avg}}$.
• (b) Find all values of $c \in [0, 3]$ that satisfy the Mean Value Theorem for Integrals.

Step 1: Compute the definite integral.

$$\int_0^3 (3x^2 - 2x + 1) \, dx = \left[ x^3 - x^2 + x \right]_0^3 = \left( (3)^3 - (3)^2 + 3 \right) - 0 = (27 - 9 + 3) = 21$$

Step 2: Calculate $f_{\text{avg}}$.

$$f_{\text{avg}} = \frac{1}{3 - 0} \int_0^3 f(x) \, dx = \frac{1}{3} (21) = 7$$

Step 3: Set $f(c) = 7$ and solve for $c$.

$$3c^2 - 2c + 1 = 7 \implies 3c^2 - 2c - 6 = 0$$

Apply the quadratic formula:

$$c = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-6)}}{2(3)} = \frac{2 \pm \sqrt{4 + 72}}{6} = \frac{2 \pm \sqrt{76}}{6} = \frac{2 \pm 2\sqrt{19}}{6} = \frac{1 \pm \sqrt{19}}{3}$$

Step 4: Check domain constraints $c \in [0, 3]$.

• $c_1 = \frac{1 + \sqrt{19}}{3} \approx \frac{1 + 4.3589}{3} \approx 1.786 \in [0, 3]$ (Valid)
• $c_2 = \frac{1 - \sqrt{19}}{3} \approx \frac{1 - 4.3589}{3} \approx -1.120 \notin [0, 3]$ (Discard)

$$\text{Conclusion: } f_{\text{avg}} = 7 \quad \text{and} \quad c = \frac{1 + \sqrt{19}}{3}$$

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Worked Example 2: Trigonometric Function

Problem: Find the average value of $f(x) = \sin x$ on $[0, \pi]$ and determine the values of $c$ where $f(c) = f_{\text{avg}}$.

Step 1: Compute the integral.

$$\int_0^\pi \sin x \, dx = \left[ -\cos x \right]_0^\pi = -\cos(\pi) - (-\cos(0)) = -(-1) - (-1) = 1 + 1 = 2$$

Step 2: Compute $f_{\text{avg}}$.

$$f_{\text{avg}} = \frac{1}{\pi - 0} (2) = \frac{2}{\pi} \approx 0.6366$$

Step 3: Solve for $c$.

$$\sin c = \frac{2}{\pi} \implies c_1 = \arcsin\left(\frac{2}{\pi}\right) \approx 0.6901 \text{ radians}$$ $$c_2 = \pi - \arcsin\left(\frac{2}{\pi}\right) \approx 3.14159 - 0.6901 \approx 2.4515 \text{ radians}$$

Both $c_1, c_2 \in [0, \pi]$, demonstrating that continuous functions can achieve their average value at multiple points on an interval.

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6. Real-World Applications: Kinematics, Thermodynamics & Economics

1. Average Velocity in Kinematics

If $v(t) = s'(t)$ represents the instantaneous velocity of an object moving along a line, its average velocity over $[t_1, t_2]$ is:

$$v_{\text{avg}} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$$

This unifies the integral definition of average value with the algebra difference quotient $\frac{\Delta s}{\Delta t}$.

2. Daily Temperature Cycles

If ambient temperature in degrees Fahrenheit follows $T(t) = 50 + 15\sin\left(\frac{\pi t}{12}\right)$ over a 24-hour cycle $t \in [0, 24]$:

$$T_{\text{avg}} = \frac{1}{24} \int_0^{24} \left( 50 + 15\sin\left(\frac{\pi t}{12}\right) \right) dt = \frac{1}{24} \left[ 50t - \frac{180}{\pi}\cos\left(\frac{\pi t}{12}\right) \right]_0^{24} = 50^\circ\text{F}$$

3. Economic Inventory Holding

If inventory level at time $t$ is $I(t) = 1000 - 50t$ over a 20-day cycle, the average inventory held is:

$$I_{\text{avg}} = \frac{1}{20} \int_0^{20} (1000 - 50t) \, dt = \frac{1}{20} \left[ 1000t - 25t^2 \right]_0^{20} = \frac{1}{20} (20000 - 10000) = 500 \text{ units}$$

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7. Comparison: Mean Value Theorems in Calculus

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8. Common CLEP Pitfalls & Traps

1. Forgetting the Normalizing Factor $\frac{1}{b - a}$: Computing only $\int_a^b f(x) \, dx$ gives the total accumulated area, not the average value. Always divide by the interval length $(b - a)$.
2. Confusing Function Average with Arithmetic Average of Endpoints: Writing $f_{\text{avg}} = \frac{f(a) + f(b)}{2}$ is ONLY valid for linear functions ($f(x) = mx + k$). For non-linear curves (quadratics, exponentials, trig), it produces severe errors.
3. Failing to Filter Extraneous $c$ Values: Always check that your algebraic solution for $c$ satisfies $a \le c \le b$. Selecting a root outside the interval is a common multiple-choice trap.
4. Discontinuous Functions: The MVT for Integrals requires continuity on $[a, b]$. If $f(x)$ has a jump or vertical asymptote, the existence of $c$ where $f(c) = f_{\text{avg}}$ is not guaranteed.