Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 5 - Integration

Problema 5.4.59b

Capitolo 5, Problema 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)

Guida verificata passo dopo passo

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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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.

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Fundamental Theorem of Calculus Part 1

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.

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Fundamental Theorem of Calculus Part 1

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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Properties of Functions

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