Skip to main content
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 5.2.53c
Capitolo 5, Problema 5.2.53c

Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.
(c) ∫₃⁶ (3ƒ(𝓍) ― g(𝓍)) d𝓍

Guida verificata passo dopo passo
1
Step 1: Recognize that the integral ∫₃⁶ (3ƒ(𝓍) ― g(𝓍)) d𝓍 can be split into separate integrals using the linearity property of integrals. This property states that ∫ₐᵇ [c₁f(𝓍) + c₂g(𝓍)] d𝓍 = c₁∫ₐᵇ f(𝓍) d𝓍 + c₂∫ₐᵇ g(𝓍) d𝓍.
Step 2: Apply the linearity property to rewrite the integral as ∫₃⁶ (3ƒ(𝓍)) d𝓍 ― ∫₃⁶ g(𝓍) d𝓍.
Step 3: Factor out the constant 3 from the first integral using the constant multiple rule, which states that ∫ₐᵇ c·f(𝓍) d𝓍 = c·∫ₐᵇ f(𝓍) d𝓍. This gives 3∫₃⁶ ƒ(𝓍) d𝓍 ― ∫₃⁶ g(𝓍) d𝓍.
Step 4: Substitute the given values for the integrals. From the problem, ∫₃⁶ ƒ(𝓍) d𝓍 = ―5 and ∫₃⁶ g(𝓍) d𝓍 = 1.
Step 5: Combine the results algebraically to evaluate the expression. The final result will be 3(―5) ― 1.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Properties of Definite Integrals

Definite integrals have several key properties, including linearity, which states that the integral of a sum is the sum of the integrals, and the integral of a constant multiplied by a function is the constant multiplied by the integral of the function. This allows us to break down complex integrals into simpler parts, making calculations more manageable.
Video consigliato:
Percorso guidato
05:43
Definition of the Definite Integral

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, stating that if a function is continuous on an interval, then the integral of its derivative over that interval equals the change in the function's values at the endpoints. This theorem is essential for evaluating definite integrals and understanding the relationship between a function and its integral.
Video consigliato:
Percorso guidato
06:11
Fundamental Theorem of Calculus Part 1

Integration of Piecewise Functions

When dealing with piecewise functions, integrals can be evaluated over different intervals where the function may change. This requires calculating the integral separately over each interval and then summing the results. Understanding how to handle piecewise functions is crucial for accurately evaluating integrals that involve different expressions across specified ranges.
Video consigliato:
Percorso guidato
05:36
Piecewise Functions
Pratica correlata
Domanda del libro di testo

Sigma notation Evaluate the following expressions.                                                                                                                                          

(c)     4                                                                                                                                                                               

       ∑ κ²                                                                                                                                                                          

       κ=1                         

Domanda del libro di testo

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.


(c) ∫₄⁰ 6𝓍(4 ― 𝓍) d(𝓍)

57
views
Domanda del libro di testo

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

(c) For an increasing or decreasing nonconstant function on an interval [a,b] and a given value of n, the value of the midpoint Riemann sum always lies between the values of the left and right Riemann sums.

63
views
Domanda del libro di testo

Sigma notation Express the following sums using sigma notation. (Answers are not unique.)

(c) 1² + 2² + 3² + 4²

80
views
Domanda del libro di testo

Approximating areas Estimate the area of the region bounded by the graph of ƒ(𝓍) = x² + 2 and the x-axis on [0, 2] in the following ways.

(c) Divide [0, 2] into n = 4 subintervals and approximate the area of the region using a right Riemann sum. Illustrate the solution geometrically.

74
views
Domanda del libro di testo

{Use of Tech} Approximating definite integrals Complete the following steps for the given integral and the given value of n. 

(c) Calculate the left and right Riemann sums for the given value of n.


∫₀^π/2 cos 𝓍 d𝓍 ; n = 4

94
views