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Mean Value Theorem for Integrals — Formula & Examples

Mean Value Theorem for Integrals

A variation of the mean value theorem which guarantees that a continuous function has at least one point where the function equals the average value of the function.

 

Mean Value Theorem for Integrals: if f is continuous on [a,b], there exists c in [a,b] such that f(c) = 1/(b−a) ∫f(x)dx

Graph showing y=f(x) curve over [a,b] with shaded area under curve equal to rectangle of height f(c), where c is between a and b.

 

See also

Theorem

Key Formula

f(c)=1b−a∫abf(x) dxf(c) = \frac{1}{b - a} \int_a^b f(x)\, dx
Where:
  • ff = A continuous function on the closed interval [a, b]
  • aa = The left endpoint of the interval
  • bb = The right endpoint of the interval
  • cc = A point in [a, b] where f(c) equals the average value of f
  • 1b−a∫abf(x) dx\frac{1}{b-a}\int_a^b f(x)\,dx = The average value of f on [a, b]

Worked Example

Problem: Let f(x) = x² on the interval [0, 3]. Find a value c guaranteed by the Mean Value Theorem for Integrals where f(c) equals the average value of f.
Step 1: Compute the definite integral of f(x) = x² from 0 to 3.
∫03x2 dx=[x33]03=273−0=9\int_0^3 x^2\, dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9
Step 2: Find the average value of f on [0, 3] by dividing the integral by the length of the interval.
Average value=13−0⋅9=3\text{Average value} = \frac{1}{3 - 0} \cdot 9 = 3
Step 3: Set f(c) equal to the average value and solve for c.
c2=3  ⟹  c=3≈1.732c^2 = 3 \implies c = \sqrt{3} \approx 1.732
Step 4: Verify that c = √3 lies in the interval [0, 3]. Since 0 ≤ 1.732 ≤ 3, the theorem is satisfied.
c=3∈[0,3]✓c = \sqrt{3} \in [0, 3] \checkmark
Answer: The value c = √3 ≈ 1.732 satisfies the Mean Value Theorem for Integrals. At this point, f(√3) = 3, which equals the average value of f on [0, 3].

Another Example

This example uses a trigonometric function and demonstrates that the theorem guarantees at least one value of c, but there can be more than one. It also involves an inverse trig solution rather than a simple algebraic one.

Problem: Let f(x) = sin(x) on the interval [0, π]. Find the value c guaranteed by the Mean Value Theorem for Integrals.
Step 1: Compute the definite integral of sin(x) from 0 to π.
∫0πsin⁡(x) dx=[−cos⁡(x)]0π=−cos⁡(π)+cos⁡(0)=1+1=2\int_0^{\pi} \sin(x)\, dx = \left[-\cos(x)\right]_0^{\pi} = -\cos(\pi) + \cos(0) = 1 + 1 = 2
Step 2: Calculate the average value of f on [0, π].
Average value=1π−0⋅2=2π≈0.6366\text{Average value} = \frac{1}{\pi - 0} \cdot 2 = \frac{2}{\pi} \approx 0.6366
Step 3: Set f(c) = 2/π and solve for c using the inverse sine function.
sin⁡(c)=2π  ⟹  c=arcsin⁡ ⁣(2π)≈0.6901\sin(c) = \frac{2}{\pi} \implies c = \arcsin\!\left(\frac{2}{\pi}\right) \approx 0.6901
Step 4: Check that c lies in [0, π]. Since 0 ≤ 0.6901 ≤ π, the condition holds. Note there is actually a second solution c = π − 0.6901 ≈ 2.4515 also in the interval, showing that more than one c can exist.
c1≈0.690,c2=π−c1≈2.452c_1 \approx 0.690, \quad c_2 = \pi - c_1 \approx 2.452
Answer: Two values satisfy the theorem: c ≈ 0.690 and c ≈ 2.452. Both lie in [0, π] and give sin(c) = 2/π.

Frequently Asked Questions

What is the difference between the Mean Value Theorem and the Mean Value Theorem for Integrals?
The standard Mean Value Theorem (for derivatives) says there exists a point c where the instantaneous rate of change f'(c) equals the average rate of change over the interval. The Mean Value Theorem for Integrals says there exists a point c where the function value f(c) equals the average value of the function over the interval. One deals with slopes (derivatives), while the other deals with function values (integrals).
Why does the function need to be continuous for the Mean Value Theorem for Integrals?
Continuity on [a, b] ensures that f attains every value between its minimum and maximum on the interval (by the Intermediate Value Theorem). Since the average value of f must lie between the minimum and maximum, continuity guarantees that f actually equals this average value at some point c. Without continuity, the function could "jump over" its average value.
What does the Mean Value Theorem for Integrals mean geometrically?
Geometrically, the area under the curve y = f(x) from a to b equals the area of a rectangle with width (b − a) and height f(c). The theorem guarantees you can always find at least one height f(c) that produces a rectangle with the same area as the region under the curve.

Mean Value Theorem for Integrals vs. Mean Value Theorem (for Derivatives)

Mean Value Theorem for IntegralsMean Value Theorem (for Derivatives)
StatementThere exists c in [a, b] where f(c) equals the average value of fThere exists c in (a, b) where f'(c) equals the average rate of change
Formulaf(c) = (1/(b−a)) ∫ₐᵇ f(x) dxf'(c) = (f(b) − f(a)) / (b − a)
Hypothesesf continuous on [a, b]f continuous on [a, b] and differentiable on (a, b)
What it findsA point where the function value equals its averageA point where the derivative equals the average slope
Geometric meaningA rectangle with equal area to the region under the curveA tangent line parallel to the secant line

Why It Matters

The Mean Value Theorem for Integrals appears throughout AP Calculus AB/BC courses and is a standard topic on the AP exam. It provides the theoretical foundation for the concept of average value of a function, which has practical applications in physics (e.g., average velocity, average temperature) and engineering. Understanding this theorem also strengthens your grasp of how the Fundamental Theorem of Calculus connects derivatives and integrals.

Common Mistakes

Mistake: Forgetting to divide by (b − a) when computing the average value, and instead treating the integral itself as the average.
Correction: The integral ∫ₐᵇ f(x) dx gives the total accumulated value (area). You must divide by the interval length (b − a) to get the average value: (1/(b−a)) ∫ₐᵇ f(x) dx.
Mistake: Assuming there is always exactly one value of c that satisfies the theorem.
Correction: The theorem guarantees at least one such c, but there can be multiple values. For example, sin(x) on [0, π] has two points where it equals its average value. Always check for all solutions within the interval.

Related Terms