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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243당신이 사용하는 게 아니라요?교과서 변경
5장, 문제 5.1.39d

Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.


{Use of Tech} ƒ(𝓍) = √x on [1,3] ; n = 4


(d) Calculate the midpoint Riemann sum.

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Step 1: Understand the problem. The goal is to calculate the midpoint Riemann sum for the function ƒ(𝓍) = √x over the interval [1, 3] with n = 4 subintervals. A midpoint Riemann sum approximates the area under the curve by summing the areas of rectangles whose heights are determined by the function value at the midpoint of each subinterval.
Step 2: Divide the interval [1, 3] into n = 4 equal subintervals. To do this, calculate the width of each subinterval, Δx, using the formula Δx = (b - a) / n, where a = 1 and b = 3. This gives Δx = (3 - 1) / 4 = 0.5.
Step 3: Determine the midpoints of each subinterval. The subintervals are [1, 1.5], [1.5, 2], [2, 2.5], and [2.5, 3]. The midpoints are calculated as the average of the endpoints of each subinterval: (1 + 1.5)/2 = 1.25, (1.5 + 2)/2 = 1.75, (2 + 2.5)/2 = 2.25, and (2.5 + 3)/2 = 2.75.
Step 4: Evaluate the function ƒ(𝓍) = √x at each midpoint. This means calculating √1.25, √1.75, √2.25, and √2.75. These values represent the heights of the rectangles for the Riemann sum.
Step 5: Multiply each function value by the width of the subinterval, Δx = 0.5, and sum the results to approximate the area under the curve. The midpoint Riemann sum is given by the formula: S = Δx * [ƒ(1.25) + ƒ(1.75) + ƒ(2.25) + ƒ(2.75)].

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Riemann Sums

Riemann sums are a method for approximating the definite integral of a function over a specified interval. They involve dividing the interval into smaller subintervals, calculating the function's value at specific points within these subintervals, and summing the products of these values and the widths of the subintervals. The midpoint Riemann sum specifically uses the midpoint of each subinterval to evaluate the function, providing a more accurate approximation than using the endpoints.
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가이드 코스
06:11
Introduction to Riemann Sums

Midpoint Rule

The midpoint rule is a specific type of Riemann sum that estimates the area under a curve by taking the value of the function at the midpoint of each subinterval. For an interval [a, b] divided into n equal parts, the midpoint of each subinterval is calculated, and the function is evaluated at these midpoints. The sum of these values, multiplied by the width of the subintervals, gives the midpoint Riemann sum, which serves as an approximation of the integral.
추천 영상:
5:50
Power Rules

Definite Integral

The definite integral of a function over an interval [a, b] represents the net area under the curve of the function between the two points a and b. It is a fundamental concept in calculus, linking the concepts of area and accumulation. The definite integral can be computed using various methods, including Riemann sums, and is denoted as ∫_a^b f(x) dx, where f(x) is the function being integrated.
추천 영상:
가이드 코스
05:43
Definition of the Definite Integral
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교과서 질문

Left and right Riemann sums Complete the following steps for the given function, interval, and value of n.

f(x) = x + 1 on [0,4]; n = 4

(d) Calculate the left and right Riemann sums.                                                                                                                                                

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(d) Assuming the velocity remains 10 m/s, for t ≥ 5, find the function that gives the displacement between t = 0 and any time t ≥ 5.

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교과서 질문

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 (d) ∫₀^π/¹⁶ sec ² 4𝓍 d𝓍

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교과서 질문

Left and right Riemann sums Complete the following steps for the given function, interval, and value of n.

ƒ(𝓍) = x² ─ 1 on [2,4]; n = 4

(d) Calculate the left and right Riemann sums. 

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교과서 질문

Left and right Riemann sums Complete the following steps for the given function, interval, and value of n.

{Use of Tech} ƒ(𝓍) = cos 𝓍 on [0. π/2]; n = 4

(d) Calculate the left and right Riemann sums.

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