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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.2.75a
Capitolo 5, Problema 5.2.75a

{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.
(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.


∫₁⁴ 2√𝓍 d𝓍

Guida verificata passo dopo passo
1
Step 1: Understand the problem. A midpoint Riemann sum approximates the value of a definite integral by dividing the interval into n subintervals, calculating the function value at the midpoint of each subinterval, and summing the areas of the rectangles formed.
Step 2: Define the interval and subintervals. The integral ∫₁⁴ 2√𝓍 d𝓍 is over the interval [1, 4]. Divide this interval into n subintervals of equal width Δ𝓍 = (4 - 1)/n = 3/n.
Step 3: Determine the midpoints of the subintervals. The midpoint of the i-th subinterval is given by 𝓍ᵢ = 1 + (i - 0.5)Δ𝓍, where i ranges from 1 to n.
Step 4: Write the function value at the midpoint. The function being integrated is f(𝓍) = 2√𝓍. At the midpoint 𝓍ᵢ, the function value is f(𝓍ᵢ) = 2√(1 + (i - 0.5)Δ𝓍).
Step 5: Express the midpoint Riemann sum in sigma notation. The sum is approximated as Sₙ = Σ (from i=1 to n) [f(𝓍ᵢ) * Δ𝓍], which becomes Sₙ = Σ (from i=1 to n) [2√(1 + (i - 0.5)(3/n)) * (3/n)].

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Midpoint Riemann Sum

A Midpoint Riemann Sum is a method for approximating the value of a definite integral. It involves dividing the interval into 'n' subintervals, calculating the midpoint of each subinterval, and then evaluating the function at these midpoints. The sum of these values, multiplied by the width of the subintervals, provides an estimate of the area under the curve.
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Left, Right, & Midpoint Riemann Sums

Sigma Notation

Sigma notation is a concise way to represent the sum of a sequence of terms. It uses the Greek letter sigma (Σ) to indicate summation, along with an index of summation that specifies the starting and ending values. In the context of Riemann sums, sigma notation allows us to express the sum of function values at midpoints over all subintervals in a clear and compact form.
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Sigma Notation

Definite Integral

A definite integral represents the signed area under a curve defined by a function over a specific interval [a, b]. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. The definite integral provides a precise value that corresponds to the accumulation of quantities, such as area, over the interval, and is often evaluated using techniques like Riemann sums.
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Definition of the Definite Integral
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{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.

(a) Write the midpoint Riemann sum in sigma notation for an arbitrary value of n.


∫₀⁴ (4𝓍― 𝓍²) d𝓍

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Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.



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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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Working with area functions Consider the function ƒ and the points a, b, and c.

(a) Find the area function A (𝓍) = ∫ₐˣ ƒ(t) dt using the Fundamental Theorem.

ƒ(𝓍) = cos 𝓍 ; a = 0 , b = π/2 , c = π

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Sigma notation Evaluate the following expressions.

(a)    10                                                                                                                                                                               

       ∑ κ                                                                                                                                                                          

       κ=1                         

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Approximating displacement The velocity in ft/s of an object moving along a line is given by v = 3t² + 1 on the interval 0 ≤ t ≤ 4, where t is measured in seconds.

(a) Divide the interval [0,4] into n = 4 subintervals, [0,1] , [1.2] , [2,3] , and [3,4]. On each subinterval, assume the object moves at a constant velocity equal to v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0, 4] (see part (a) of the figure)

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