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Home/Calculus/Chapter 6/Mean Value Theorems

CALC-6.5

3-4 hours

Advanced Topic

Mean Value Theorems for Integrals

Powerful theorems that express integrals in terms of function values at specific points: the First MVT, Bonnet's theorem, and the Weierstrass form.

Learning Objectives

Prove the first mean value theorem for integrals

Apply the weighted first mean value theorem

Understand the second mean value theorem (Bonnet form)

Master the Weierstrass form of the second MVT

Apply Abel's transformation technique

Solve limit problems using mean value theorems

Connect integral MVT to differential MVT

1. Motivation: Averaging Function Values

The Central Question

Given a continuous function $f$ on $[a, b]$ , can we find a point $\xi$ where the function value equals the “average value” of $f$ over the interval?

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

The Mean Value Theorem for integrals guarantees such a point exists!

Remark 6.10: Geometric Interpretation

The area under $f$ from $a$ to $b$ equals the area of a rectangle with:

Width: $b - a$
Height: $f(\xi)$ (the mean value)

\int_a^b f(x)\,dx = f(\xi) \cdot (b - a)

Area under curve = Area of rectangle

2. First Mean Value Theorem

Theorem 6.28: First Mean Value Theorem for Integrals

Let $g \in R[a, b]$ with $g(x)$ not changing sign (i.e., $g \geq 0$ or $g \leq 0$ on $[a, b]$ ).

If $f \in R[a, b]$ with $m \leq f(x) \leq M$ , then there exists $\mu \in [m, M]$ such that:

\int_a^b f(x)g(x)\,dx = \mu \int_a^b g(x)\,dx

If additionally $f \in C[a, b]$ , then $\exists \xi \in [a, b]$ with $\mu = f(\xi)$ :

\int_a^b f(x)g(x)\,dx = f(\xi)\int_a^b g(x)\,dx

Proof of Theorem 6.28:

Assume $g \geq 0$ . Since $m \leq f(x) \leq M$ :

mg(x) \leq f(x)g(x) \leq Mg(x)

Integrating:

m\int_a^b g(x)\,dx \leq \int_a^b f(x)g(x)\,dx \leq M\int_a^b g(x)\,dx

If $\int g = 0$ , then $\int fg = 0$ and any $\mu$ works.

If $\int g > 0$ , define:

\mu = \frac{\int_a^b f(x)g(x)\,dx}{\int_a^b g(x)\,dx}

Then $m \leq \mu \leq M$ .

If $f$ is continuous, by IVT, $f(\xi) = \mu$ for some $\xi \in [a, b]$ .

∎

Corollary 6.8: Basic Form (g ≡ 1)

If $f \in C[a, b]$ , then $\exists \xi \in [a, b]$ :

\int_a^b f(x)\,dx = f(\xi)(b - a)

Example 6.36: Finding the Mean Value Point

Problem: For $f(x) = x^2$ on $[0, 3]$ , find $\xi$ from the First MVT.

Solution:

\int_0^3 x^2\,dx = \frac{x^3}{3}\Big|_0^3 = 9

By First MVT: $9 = \xi^2 \cdot 3$ , so $\xi^2 = 3$ .

Therefore: $\xi = \sqrt{3} \approx 1.732 \in [0, 3]$

Example 6.37: Application to Limits

Evaluate: $\lim_{n \to \infty} \int_{a/n}^{b/n} \frac{f(x)}{x}\,dx$ where $f \in C[0, c]$ with $0 < a < b \leq c$ .

Solution:

By First MVT with $g(x) = 1/x$ :

\int_{a/n}^{b/n} \frac{f(x)}{x}\,dx = f(\xi_n) \int_{a/n}^{b/n} \frac{dx}{x} = f(\xi_n) \ln\frac{b}{a}

where $\xi_n \in [a/n, b/n]$ .

As $n \to \infty$ : $\xi_n \to 0$ , so $f(\xi_n) \to f(0)$ .

Result: $\lim = f(0) \ln(b/a)$

3. Second Mean Value Theorem (Bonnet)

Theorem 6.29: Bonnet's Theorem

Let $g \in R[a, b]$ .

(1) If $f$ is non-negative and decreasing on $[a, b]$ :

\int_a^b f(x)g(x)\,dx = f(a)\int_a^\xi g(x)\,dx

for some $\xi \in [a, b]$ .

(2) If $f$ is non-negative and increasing on $[a, b]$ :

\int_a^b f(x)g(x)\,dx = f(b)\int_\xi^b g(x)\,dx

for some $\xi \in [a, b]$ .

Proof of Theorem 6.29 (Case 1):

Let $G(x) = \int_a^x g(t)\,dt$ . We'll use integration by parts in the generalized sense.

Assume first $g \in C[a,b]$ and $f \in C^1[a,b]$ . Then:

\int_a^b f(x)g(x)\,dx = \int_a^b f(x)\,dG(x) = f(x)G(x)\Big|_a^b - \int_a^b G(x)f'(x)\,dx

Since $G(a) = 0$ :

= f(b)G(b) - \int_a^b G(x)f'(x)\,dx

Since $f' \leq 0$ (decreasing) and $m \leq G(x) \leq M$ :

mf(b) - m[f(b) - f(a)] \leq \int_a^b fg \leq Mf(b) - M[f(b) - f(a)]

Simplifying: $mf(a) \leq \int fg \leq Mf(a)$ .

By IVT for $G$ , there exists $\xi$ with $G(\xi) = \int_a^\xi g = \int fg / f(a)$ .

∎

Example 6.38: Applying Bonnet's Theorem

Problem: If $f \in C^2[-1, 1]$ and $f(0) = 0$ , prove $\exists \xi \in [-1, 1]$ :

f''(\xi) = 3\int_{-1}^1 f(x)\,dx

Solution Sketch:

Let $g(x) = 1 - x^2$ . Note $g \geq 0$ on $[-1, 1]$ and $g(-1) = g(1) = 0$ .

Using integration by parts twice and applying the First MVT to the resulting integral gives the result.

4. Weierstrass Form

Theorem 6.30: Weierstrass Form of Second MVT

Let $f$ be monotonic on $[a, b]$ and $g \in R[a, b]$ . Then $\exists \xi \in [a, b]$ :

\int_a^b f(x)g(x)\,dx = f(a)\int_a^\xi g(x)\,dx + f(b)\int_\xi^b g(x)\,dx

Proof of Theorem 6.30:

Write $f = f(a) + [f - f(a)]$ if increasing, or $f = f(b) + [f - f(b)]$ if decreasing.

Apply Bonnet's theorem to the shifted function (which is now non-negative monotone).

The details involve careful bookkeeping but follow the same pattern as Bonnet's proof.

∎

Example 6.39: Using Weierstrass Form

Estimate: $\int_0^\pi x\sin x\,dx$ using Second MVT.

Solution:

Here $f(x) = x$ (increasing), $g(x) = \sin x$ .

By Weierstrass: $\exists \xi \in [0, \pi]$ :

\int_0^\pi x\sin x\,dx = 0 \cdot \int_0^\xi \sin x\,dx + \pi \int_\xi^\pi \sin x\,dx

= \pi[1 + \cos\xi]

Actual value: $\int_0^\pi x\sin x\,dx = \pi$ (by parts).

So $1 + \cos\xi = 1$ , giving $\xi = \pi/2$ .

5. Abel's Transformation

Definition 6.12: Abel's Lemma (Summation by Parts)

For sequences $\{a_k\}$ and $\{b_k\}$ , with partial sums $A_k = \sum_{i=1}^k a_i$ :

\sum_{k=1}^n a_k b_k = A_n b_n - \sum_{k=1}^{n-1} A_k(b_{k+1} - b_k)

This is the discrete analog of integration by parts: $\int u\,dv = uv - \int v\,du$ .

Theorem 6.31: Abel's Lemma for Integrals

For $f, g \in R[a, b]$ with $F(x) = \int_a^x f(t)\,dt$ and $G(x) = \int_a^x g(t)\,dt$ :

\int_a^b F(x)g(x)\,dx = F(x)G(x)\Big|_a^b - \int_a^b G(x)f(x)\,dx

Remark 6.11: Connection to Bonnet's Theorem

Abel's transformation is the key tool in proving Bonnet's theorem. By choosing $f$ appropriately, we can control the transformed integral and extract the desired $\xi$ .

6. Applications

Example 6.40: Limit Evaluation

Prove: $\lim_{n \to \infty} \int_{a/n}^{b/n} \frac{f(x)}{x}\,dx = f(0)\ln\frac{b}{a}$ for $f \in C[0, c]$ .

Solution:

By First MVT: $\int_{a/n}^{b/n} \frac{f(x)}{x}\,dx = f(\xi_n)\ln\frac{b}{a}$

where $\xi_n \in [a/n, b/n]$ .

As $n \to \infty$ : $\xi_n \to 0$ , so by continuity $f(\xi_n) \to f(0)$ .

Example 6.41: Existence of Zeros

Prove: If $f \in C[0, \pi]$ , $\int_0^\pi f(x)\,dx = 0$ , and $\int_0^\pi f(x)\cos x\,dx = 0$ , then $f$ has at least two zeros in $(0, \pi)$ .

Solution Sketch:

From the first condition: $F(\pi) = 0$ where $F(x) = \int_0^x f(t)\,dt$ .

By Rolle's theorem on $F$ : $\exists \xi_1$ with $F'(\xi_1) = f(\xi_1) = 0$ .

The second condition, combined with integration by parts and the First MVT, yields another zero.

Example 6.42: Oscillating Integrals

Prove: $\int_a^b f(x)\sin(\lambda x)\,dx \to 0$ as $\lambda \to \infty$ (Riemann-Lebesgue).

Proof Sketch:

For $f$ monotone, use Second MVT:

\int_a^b f(x)\sin(\lambda x)\,dx = f(a)\int_a^\xi \sin(\lambda x)\,dx + f(b)\int_\xi^b \sin(\lambda x)\,dx

Each integral on the right is $O(1/\lambda)$ .

For general $f \in R[a,b]$ , approximate by step functions.

7. Common Mistakes

❌ Assuming ξ is unique

The MVT only guarantees existence, not uniqueness of ξ!

❌ Forgetting continuity for ξ

Without continuity, we only get μ ∈ [m, M], not f(ξ) = μ.

❌ Wrong direction in Bonnet

For decreasing f: use f(a). For increasing f: use f(b)!

❌ Ignoring sign conditions

Bonnet requires f ≥ 0. For general monotone f, use Weierstrass.

8. Additional Examples

Example 6.43: Integral with Oscillating Weight

Problem: Show $\int_0^\pi \sin(nx)\sin(\lambda x)\,dx \to 0$ as $\lambda \to \infty$ .

Solution:

Use product-to-sum: $\sin(nx)\sin(\lambda x) = \frac{1}{2}[\cos((n-\lambda)x) - \cos((n+\lambda)x)]$

\int_0^\pi \sin(nx)\sin(\lambda x)\,dx = \frac{1}{2}\left[\frac{\sin((n-\lambda)\pi)}{n-\lambda} - \frac{\sin((n+\lambda)\pi)}{n+\lambda}\right]

Both terms are $O(1/\lambda)$ as $\lambda \to \infty$ .

Example 6.44: Average Value Problem

Problem: Find the average value of $f(x) = \sin^2 x$ on $[0, 2\pi]$ .

Solution:

\bar{f} = \frac{1}{2\pi}\int_0^{2\pi} \sin^2 x\,dx = \frac{1}{2\pi}\int_0^{2\pi} \frac{1 - \cos 2x}{2}\,dx

= \frac{1}{4\pi}\left[x - \frac{\sin 2x}{2}\right]_0^{2\pi} = \frac{1}{4\pi} \cdot 2\pi = \frac{1}{2}

The mean value is $1/2$ , achieved at $\xi = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4$ .

Example 6.45: Integral Inequality via MVT

Problem: If $f \in C[0, 1]$ with $\int_0^1 f = 0$ , prove $\int_0^1 xf(x)\,dx \leq \frac{1}{2}\max|f|$ .

Solution:

Let $M = \max_{[0,1]}|f|$ . Since $\int_0^1 f = 0$ :

\left|\int_0^1 xf(x)\,dx\right| = \left|\int_0^1 xf(x)\,dx - 0 \cdot \int_0^1 f\right| = \left|\int_0^1 xf(x)\,dx\right|

Using the weighted MVT with weight $x$ :

\left|\int_0^1 xf(x)\,dx\right| \leq M\int_0^1 x\,dx = \frac{M}{2}

Example 6.46: Convergence of Integral Sequences

Problem: For $f \in C[0, 1]$ , find $\lim_{n \to \infty} n\int_0^1 x^n f(x)\,dx$ .

Solution:

By First MVT: $\int_0^1 x^n f(x)\,dx = f(\xi_n)\int_0^1 x^n\,dx = \frac{f(\xi_n)}{n+1}$

where $\xi_n \in [0, 1]$ .

So: $n\int_0^1 x^n f(x)\,dx = \frac{n}{n+1}f(\xi_n)$

Key insight: The weight $x^n$ concentrates near $x = 1$ as $n \to \infty$ .

More carefully: $\xi_n \to 1$ as $n \to \infty$ .

Result: $\lim = f(1)$

Example 6.47: Double Integral Connection

Problem: Using integral MVT, prove Cauchy-Schwarz: $(\int fg)^2 \leq \int f^2 \cdot \int g^2$ .

Proof Sketch:

Consider $\int_a^b (f(x) + tg(x))^2\,dx \geq 0$ for all $t \in \mathbb{R}$ .

\int f^2 + 2t\int fg + t^2\int g^2 \geq 0

This is a non-negative quadratic in $t$ , so discriminant ≤ 0:

4(\int fg)^2 - 4\int f^2 \cdot \int g^2 \leq 0

Example 6.48: Bonnet Applied to Bessel Functions

Problem: Estimate $\int_0^\pi \frac{\cos(nx)}{1+x}\,dx$ for large $n$ .

Solution:

Let $f(x) = \frac{1}{1+x}$ (decreasing, positive) and $g(x) = \cos(nx)$ .

By Bonnet: $\exists \xi \in [0, \pi]$ :

\int_0^\pi \frac{\cos(nx)}{1+x}\,dx = f(0)\int_0^\xi \cos(nx)\,dx = \frac{\sin(n\xi)}{n}

Since $|\sin(n\xi)| \leq 1$ : $\left|\int\right| \leq \frac{1}{n} \to 0$ .

9. Practice Problems

Problem Set A: First Mean Value Theorem

1. Find $\xi$ for $f(x) = e^x$ on $[0, 1]$ .

2. Find average value of $\cos x$ on $[0, \pi]$ .

3. Evaluate $\lim_{h \to 0} \frac{1}{h}\int_a^{a+h} f(x)\,dx$ .

4. If $\int_0^1 f = \int_0^1 g$ , prove $f(\xi) = g(\eta)$ for some $\xi, \eta$ .

Problem Set B: Second Mean Value Theorem

5. Apply Bonnet to $\int_1^2 \frac{\sin x}{x}\,dx$ .

6. Show $\int_0^1 x^n\sin(\pi x)\,dx = O(1/n)$ .

7. Use Weierstrass for $\int_0^1 (1-x)\cos(nx)\,dx$ .

8. Prove: $\int_0^\infty \frac{\sin x}{x}\,dx$ converges.

Problem Set C: Applications and Limits

9. $\lim_{n \to \infty} \int_0^1 \frac{f(x)}{1 + nx^2}\,dx$

10. $\lim_{n \to \infty} \int_0^{\pi/2} \sin^n x\,dx \cdot \sqrt{n}$

11. Prove $\int_0^a f(x)\,dx = af(\theta a)$ for some $\theta \in (0,1)$ .

12. Show $\lim_{\lambda \to \infty} \int_0^1 e^{-\lambda x}f(x)\,dx = 0$ .

Remark 6.12: Answers to Selected Problems

\xi = \ln(e-1)

0

f(a)

10:

\sqrt{\pi/2}

10. Historical Context

Pierre Ossian Bonnet (1819-1892)

French mathematician who proved the second mean value theorem in its most useful form. Also known for fundamental contributions to differential geometry.

Karl Weierstrass (1815-1897)

German mathematician known as the “father of modern analysis”. His rigorous approach to calculus led to the epsilon-delta definitions we use today.

Niels Henrik Abel (1802-1829)

Norwegian mathematician who proved the impossibility of solving quintic equations by radicals. His summation technique (Abel's lemma) is a key tool in analysis.

Augustin-Louis Cauchy (1789-1857)

French mathematician who first stated the mean value theorem for integrals in its modern form, establishing the foundations of rigorous analysis.

11. Connections to Other Topics

Differential MVT

The integral MVT is closely related to the differential MVT via FTC. Both express global properties in terms of local values.

Fourier Series

The Riemann-Lebesgue lemma (proved via Second MVT) is essential for understanding Fourier coefficient decay.

Improper Integrals

The Second MVT is key for proving convergence of improper integrals with oscillating integrands.

Numerical Integration

Error bounds for numerical methods like the midpoint rule use the First MVT.

Asymptotic Analysis

MVTs help extract leading-order behavior of integrals depending on parameters.

Probability Theory

Expected values and weighted averages are generalizations of the mean value concept.

12. Theorem Summary

Theorem	Conditions	Conclusion
First MVT (basic)	$f \in C[a,b]$	$\int f = f(\xi)(b-a)$
Weighted First MVT	$f \in C, g \geq 0$	$\int fg = f(\xi)\int g$
Bonnet (↘)	$f \geq 0$ decreasing	$\int fg = f(a)\int_a^\xi g$
Bonnet (↗)	$f \geq 0$ increasing	$\int fg = f(b)\int_\xi^b g$
Weierstrass	$f$ monotone	$\int fg = f(a)\int_a^\xi g + f(b)\int_\xi^b g$

13. Detailed Proof Techniques

Proof Strategy for Second MVT

Step 1: Define Auxiliary Function

Let $G(x) = \int_a^x g(t)\,dt$ . This is the “cumulative weight” function.

Step 2: Apply Integration by Parts

Write $\int fg = \int f\,dG$ and apply Abel's transformation.

Step 3: Bound the Integral

Use monotonicity of $f$ to bound $\int G\,df$ by min/max of $G$ .

Step 4: Apply IVT

Since $G$ is continuous and the bounds are achieved, IVT gives $\xi$ .

Remark 6.13: Alternative Proof via Riemann Sums

Another approach to the Second MVT uses Riemann sums directly:

1. Partition $[a, b]$ and write the Riemann sum for $\int fg$ .

2. Use Abel's summation formula (discrete version).

3. Take the limit as mesh size → 0.

This approach is more elementary but requires careful handling of limits.

14. Challenge Problems

Competition-Level Problems

Problem 1:

If $f \in C^1[0, 1]$ with $f(0) = f(1) = 0$ , prove:

\left|\int_0^1 f(x)\,dx\right| \leq \frac{1}{4}\max_{[0,1]}|f'(x)|

Problem 2:

For $f \in C[0, 1]$ with $\int_0^1 f = 0$ and $\int_0^1 xf(x)\,dx = 0$ , prove $f$ has at least two zeros in $(0, 1)$ .

Problem 3:

Prove: $\lim_{n \to \infty} \int_0^1 \frac{n x^{n-1}}{1 + x^n}\,dx = \ln 2$

Problem 4:

If $f$ is continuous and $\int_0^x f(t)\,dt = xf(x)$ for all $x$ , prove $f(x) = cx$ for some constant $c$ .

Problem 5:

Use the Second MVT to prove the Dirichlet test for improper integrals: if $f$ is monotone → 0 and $\int_a^x g$ is bounded, then $\int_a^\infty fg$ converges.

Remark 6.14: Hints for Challenge Problems

Problem 1: Use $f(x) = \int_0^x f'(t)\,dt$ and bound carefully.

Problem 2: Consider $F(x) = \int_0^x f(t)\,dt$ and apply Rolle twice.

Problem 3: Substitute $u = x^n$ and use dominated convergence.

Problem 4: Differentiate both sides using FTC.

15. Extended Formula Reference

Average Value

\bar{f} = \frac{1}{b-a}\int_a^b f(x)\,dx

The mean value $f(\xi)$ equals $\bar{f}$ .

Weighted Average

\bar{f}_w = \frac{\int_a^b f(x)w(x)\,dx}{\int_a^b w(x)\,dx}

For weight $w \geq 0$ .

Root Mean Square

f_{\text{rms}} = \sqrt{\frac{1}{b-a}\int_a^b f^2(x)\,dx}

Used in physics and signal processing.

Generalized Mean

M_p(f) = \left(\frac{1}{b-a}\int_a^b |f|^p\,dx\right)^{1/p}

The $L^p$ norm divided by $(b-a)^{1/p}$ .

Remark 6.15: Inequalities Between Means

For non-negative continuous $f$ :

\min f \leq M_1(f) \leq M_2(f) \leq \max f

Arithmetic mean ≤ Root mean square. These inequalities follow from convexity arguments.

16. Summary: When to Use Each Theorem

Scenario	Best Tool	Why
Simple integral, find average	First MVT	Gives $f(\xi)$ directly
Product with weight $g \geq 0$	Weighted MVT	Factors out $f(\xi)$
Monotone $f \geq 0$ , oscillating $g$	Bonnet	Bounds using endpoint values
General monotone $f$	Weierstrass	No sign restriction on $f$
Limit as parameter → ∞	Either MVT + limit	Extract $\xi$ behavior
Proving integral vanishes	Second MVT	Bounds by $O(1/n)$

17. Further Reading and Extensions

Generalized MVT

The Cauchy mean value theorem generalizes to: $\frac{\int f g}{\int h g} = \frac{f(\xi)}{h(\xi)}$ under appropriate conditions.

Lebesgue Integration

Mean value theorems extend to Lebesgue integrals with measure-theoretic conditions replacing continuity/monotonicity.

Stochastic MVT

In probability theory, the MVT connects to expectations: $E[f(X)] = f(\xi)$ for some $\xi$ in the support.

Multivariable Extensions

The MVT extends to multiple integrals: $\iint_D f = f(\xi, \eta) \cdot \text{Area}(D)$ for convex domains.

Remark 6.16: Chapter Summary

This chapter covered the mean value theorems for integrals:

First MVT: Expresses integrals via function values at interior points
Weighted MVT: Handles products with non-negative weights
Bonnet's Theorem: For monotone non-negative functions
Weierstrass Form: General monotone functions, any sign
Abel's Transformation: Key proof technique (summation by parts)

These tools are essential for estimating integrals, proving convergence, and evaluating limits.

Mean Value Theorems Quiz

Questions

Correct

Accuracy

The First Mean Value Theorem states that for

f \in C[a,b]

Easy

Not attempted

The 'mean value' in the First MVT is:

Easy

Not attempted

For the weighted First MVT with

g(x) \geq 0

Medium

Not attempted

Bonnet's theorem (Second MVT, monotone decreasing

f \geq 0

) gives:

Hard

Not attempted

For monotone increasing

f

with

f \geq 0

, Bonnet gives:

Hard

Not attempted

The Weierstrass form combines both Bonnet cases for monotone

f

Hard

Not attempted

What condition on

f

is needed for the First MVT conclusion

\exists \xi

Medium

Not attempted

Abel's transformation relates to:

Medium

Not attempted

Frequently Asked Questions

How is the integral MVT related to the differential MVT?

The integral MVT for f says ∫f = f(ξ)(b-a). If F' = f, then by differential MVT, F(b)-F(a) = F'(ξ)(b-a) = f(ξ)(b-a). So the integral MVT follows from the differential MVT via FTC!

Why do we need f continuous for the First MVT?

Continuity ensures f attains its mean value. If f is just integrable, the average might not equal f(x) for any x (think of discontinuous functions).

What's special about Bonnet's theorem?

Bonnet's theorem handles products fg where f is monotone. The key insight is that the 'weight' f concentrates the integral near one endpoint.

When is the weighted MVT useful?

It's useful when computing limits of integrals or estimating integrals where one factor varies slowly (the weight g) compared to another (the function f).

How does Abel's transformation work?

Like integration by parts, but for sums. It transforms Σaₖbₖ into a sum involving partial sums Aₖ and differences Δbₖ, useful for proving convergence.

Key Takeaways

First MVT

\int_a^b f = f(\xi)(b-a)

Weighted MVT

\int_a^b fg = f(\xi)\int_a^b g

Bonnet (↘)

\int fg = f(a)\int_a^\xi g

Weierstrass

\int fg = f(a)\int_a^\xi g + f(b)\int_\xi^b g

Bonnet (↗)

\int fg = f(b)\int_\xi^b g

Abel Transform

\sum a_k b_k = A_n b_n - \sum A_k \Delta b_k

Average Value

\bar{f} = \frac{1}{b-a}\int_a^b f

Key Insight

Monotonicity of $f$ controls where integral concentrates

Quick Decision Guide

When to Use First MVT

Finding the average value of a function
Showing existence of a point with specific property
Converting integrals to function evaluations
Limit problems involving integrals

When to Use Second MVT

Products with oscillating factors (sin, cos)
Proving convergence of improper integrals
Riemann-Lebesgue lemma applications
Bounding integrals by endpoint values