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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.2.69a
Capitolo 5, Problema 5.2.69a

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
(a) If ƒ is a constant function on the interval [a,b], then the right and left Riemann sums give the exact value of ∫ₐᵇ ƒ(𝓍) d𝓍, for any positive integer n.

Guida verificata passo dopo passo
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Step 1: Begin by understanding the concept of a constant function. A constant function is one where the value of ƒ(𝓍) remains the same for all 𝓍 in the interval [a, b]. This means ƒ(𝓍) = c, where c is a constant.
Step 2: Recall the definition of the definite integral ∫ₐᵇ ƒ(𝓍) d𝓍. For a constant function, the integral simplifies to c(b - a), where c is the constant value of the function and [a, b] is the interval of integration.
Step 3: Understand the concept of Riemann sums. Both left and right Riemann sums approximate the area under the curve by dividing the interval [a, b] into n subintervals and summing up the areas of rectangles. For a constant function, the height of each rectangle is the same (equal to c), and the width of each subinterval is Δ𝓍 = (b - a)/n.
Step 4: Calculate the left and right Riemann sums for a constant function. Since the height of the rectangles is constant (c), both the left and right Riemann sums will compute the exact area under the curve as c(b - a), regardless of the number of subintervals n.
Step 5: Conclude that the statement is true. For a constant function on [a, b], both the left and right Riemann sums give the exact value of ∫ₐᵇ ƒ(𝓍) d𝓍 for any positive integer n, because the function's value does not vary across the interval, making the approximation exact.

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Riemann Sums

Riemann sums are a method for approximating the integral of a function over an interval by dividing the interval into smaller subintervals. For each subinterval, a sample point is chosen, and the function's value at that point is multiplied by the width of the subinterval. The sum of these products gives an approximation of the area under the curve, which becomes exact as the number of subintervals increases.
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Introduction to Riemann Sums

Constant Functions

A constant function is a function that always returns the same value regardless of the input. Mathematically, it can be expressed as f(x) = c, where c is a constant. In the context of integration, the area under the curve of a constant function over an interval is simply the product of the constant value and the length of the interval, making it straightforward to calculate using Riemann sums.
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Exponential Functions

Exact Value of an Integral

The exact value of an integral represents the precise area under the curve of a function over a specified interval. For a constant function, this value can be calculated using the formula ∫ₐᵇ c dx = c(b - a). Riemann sums, when applied to constant functions, yield the exact value because the function's value does not change across the interval, ensuring that both left and right sums converge to the same result.
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Average Value of a Function
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Zero net area Consider the function ƒ(𝓍) = 𝓍² ― 4𝓍 .

(a) Graph ƒ on the interval 𝓍 ≥ 0.

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Suppose ƒ is an odd function, ∫₀⁴ ƒ(𝓍) d𝓍 = 3 , and ∫₀⁸ ƒ(𝓍) d𝓍 = 9 .


(a) Evaluate ∫₋₈⁴ ƒ(𝓍) d𝓍 .

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Free fall On October 14, 2012, Felix Baumgartner stepped off a balloon capsule at an altitude of almost 39 km above Earth’s surface and began his free fall. His velocity in m/s during the fall is given in the figure. It is claimed that Felix reached the speed of sound 34 seconds into his fall and that he continued to fall at supersonic speed for 30 seconds. (Source: http://www.redbullstratos.com)

(a) Divide the interval [34, 64] into n = 5 subintervals with the gridpoints x₀ = 34 , x₁ = 40 , x₂ = 46 , x₃ = 52 , x₄ = 58 , and x₅ = 64. Use left and right Riemann sums to estimate how far Felix fell while traveling at supersonic speed.

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Area functions for the same linear function Let ƒ(t) = 2t ― 2 and consider the two area functions A (𝓍) = ∫₁ˣ ƒ(t) dt and F(𝓍) = ∫₄ˣ ƒ(t) dt .

(a) Evaluate A (2) and A (3). Then use geometry to find an expression for A (𝓍) , for 𝓍 ≥ 1 .

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Properties of integrals Use only the fact that ∫₀⁴ 3𝓍 (4 ―𝓍) d𝓍 = 32, and the definitions and properties of integrals, to evaluate the following integrals, if possible.

(a) ∫₄⁰ 3𝓍(4 ― 𝓍) d(𝓍)

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

(a) ∫¹₋₁ 𝓍ƒ(𝓍²) d𝓍

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