Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 5 - Integration

Problema 5.R.99b

Capitolo 5, Problema 5.R.99b

(b) Find the average value of ƒ shown in the figure on the interval [2,6] and then find the point(s) c in (2, 6) guaranteed to exist by the Mean Value Theorem for Integrals.

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Identify the function \( f(x) \) from the graph on the interval \([2,6]\). The graph shows a triangular shape with a peak at \( (4,5) \), increasing linearly from \( (2,1) \) to \( (4,5) \), then decreasing linearly from \( (4,5) \) to \( (6,1) \).

To find the average value of \( f \) on \([2,6]\), use the formula for the average value of a function: \[\text{Average value} = \frac{1}{6-2} \int_2^6 f(x) \, dx = \frac{1}{4} \int_2^6 f(x) \, dx.\]

Calculate the integral \( \int_2^6 f(x) \, dx \) by finding the area under the curve. Since the graph forms a triangle with base length \(6 - 2 = 4\) and height \(5 - 1 = 4\), the area is \[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4.\]

Use the area found as the value of the integral \( \int_2^6 f(x) \, dx \), then substitute it back into the average value formula to express the average value of \( f \) on \([2,6]\).

Apply the Mean Value Theorem for Integrals, which guarantees at least one point \( c \in (2,6) \) such that \[f(c) = \text{Average value of } f \text{ on } [2,6].\] Find \( c \) by solving the equation \( f(c) = \text{average value} \) using the piecewise linear definition of \( f(x) \) from the graph.

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Average Value of a Function

The average value of a function f on an interval [a, b] is given by (1/(b - a)) times the integral of f(x) from a to b. It represents the constant value that, if the function were flat, would yield the same area under the curve as the actual function over that interval.

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Average Value of a Function

Definite Integral and Area Under the Curve

The definite integral of a function over an interval [a, b] calculates the net area between the function's graph and the x-axis. This area is essential for finding the average value and understanding the accumulation of quantities represented by the function.

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Definition of the Definite Integral

Mean Value Theorem for Integrals

The Mean Value Theorem for Integrals states that for a continuous function on [a, b], there exists at least one point c in (a, b) where the function's value equals its average value over [a, b]. This guarantees a point where f(c) matches the average height of the function.

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Fundamental Theorem of Calculus Part 1

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