Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.4.59b

Chapter 5, Problem 5.4.59b

Generalizing the Mean Value Theorem for Integrals Suppose ƒ and g are continuous on [a, b] and let h(𝓍) = (𝓍―b) ∫ₐˣ ƒ(t) dt + (𝓍―a) ∫ₓᵇg(t)dt.
(b) Show that there is a number c in (a, b) such that ∫ₐᶜ ƒ(t) dt = ƒ(c) (b ― c)

(Source: The College Mathematics Journal, 33, 5, Nov 2002)

Verified step by step guidance

Start by recalling the given function: \(h(\mathcal{x}) = (\mathcal{x} - b) \int_a^{\mathcal{x}} f(t) \, dt + (\mathcal{x} - a) \int_{\mathcal{x}}^b g(t) \, dt\). We are asked to show that there exists a \(c \in (a,b)\) such that \(\int_a^c f(t) \, dt = f(c)(b - c)\).

Focus on the part of the problem involving \(f\). Consider the function \(H(x) = (x - b) \int_a^x f(t) \, dt\). Notice that this is part of \(h(x)\) and involves the integral of \(f\) from \(a\) to \(x\).

Differentiate \(H(x)\) with respect to \(x\) using the product rule and the Fundamental Theorem of Calculus. Recall that if \(F(x) = \int_a^x f(t) \, dt\), then \(F'(x) = f(x)\). So, \(H'(x) = \frac{d}{dx} \left[(x - b) F(x)\right] = F(x) + (x - b) f(x)\).

Set \(H'(c) = 0\) for some \(c \in (a,b)\) to find a critical point. This gives the equation: \(F(c) + (c - b) f(c) = 0\). Rewrite this as: \(\int_a^c f(t) \, dt = f(c)(b - c)\), which is exactly what we want to prove.

To justify the existence of such a \(c\), apply Rolle's Theorem or the Mean Value Theorem to \(H(x)\) on \([a,b]\). Since \(H(a) = H(b) = 0\), there must be some \(c \in (a,b)\) where \(H'(c) = 0\), completing the proof.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Mean Value Theorem for Integrals

This theorem states that if a function is continuous on [a, b], there exists a point c in (a, b) where the integral average equals the function value, i.e., ∫ₐᵇ f(t) dt = f(c)(b - a). It connects the average value of a function over an interval to a specific function value inside that interval.

Fundamental Theorem of Calculus

This theorem links differentiation and integration, stating that if F(x) = ∫ₐˣ f(t) dt, then F'(x) = f(x). It allows us to differentiate integral expressions and is essential for analyzing functions defined by integrals, like h(x) in the problem.

Continuity and Intermediate Value Property

Continuity of functions f and g on [a, b] ensures the integrals and constructed functions are well-behaved. The Intermediate Value Theorem guarantees the existence of points where certain equalities hold, which is crucial for proving the existence of c satisfying the given integral equation.

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Related Practice

Textbook Question

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(b) Evaluate each sum using a calculator with n = 20, 50, and 100. Use these values to estimate the value of the integral.

∫₁⁴ 2√𝓍 d𝓍

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ, ƒ', and ƒ'' are continuous functions for all real numbers.

(b) ∫ (ƒ(𝓍))ⁿ ƒ'(𝓍) d𝓍 = 1/(n + 1) (ƒ(𝓍))ⁿ⁺¹ + C , n ≠ ―1 .

Textbook Question

Substitutions Suppose ƒ is an even function with ∫₀⁸ ƒ(𝓍) d𝓍 = 9 . Evaluate each integral.

(b) ∫²₋₂ 𝓍²ƒ(𝓍³) d𝓍

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(b) If ƒ is a linear function on the interval [a,b] , then a midpoint Riemann sums give the exact value of ∫ₐᵇ ƒ(𝓍) d𝓍, for any positive integer n.

Textbook Question

Working with area functions Consider the function ƒ and its graph.

(b) Estimate the points (if any) at which A has a local maximum or minimum.

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Textbook Question

Area functions for constant functions Consider the following functions ƒ and real numbers a (see figure).

(b) Verify that .A'(𝓍) = ƒ(𝓍)

ƒ(t) = 5 , a = -5