Domanda del libro di testo
Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).
b) Verify that A'(𝓍) = ƒ(𝓍).
ƒ(t) = 4t + 2 , a = 0
47
views
Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Non è quello che usi tu?Cambia libro di testoScegli un libro di testo diverso
TOC.tableOfContents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.2.57b
Using properties of integrals Use the value of the first integral I to evaluate the two given integrals.
I = ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4
(b) ∫₁⁰ (2𝓍―𝓍³) d𝓍
Guida verificata passo dopo passo1
Step 1: Recognize that the integral given in part (b) is the reverse of the integral provided in part (I). Specifically, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 is the same as ∫₀¹ (𝓍³ ― 2𝓍) d𝓍, but with the limits of integration swapped.
Step 2: Use the property of integrals that states swapping the limits of integration changes the sign of the integral. Mathematically, ∫ₐᵇ f(𝓍) d𝓍 = -∫ᵇₐ f(𝓍) d𝓍.
Step 3: Apply this property to the given integral. Since ∫₀¹ (𝓍³ ― 2𝓍) d𝓍 = ―3/4, swapping the limits gives ∫₁⁰ (𝓍³ ― 2𝓍) d𝓍 = 3/4.
Step 4: Notice that the integrand in part (b) is written as (2𝓍 ― 𝓍³), which is the negative of (𝓍³ ― 2𝓍). Therefore, ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -∫₁⁰ (𝓍³ ― 2𝓍) d𝓍.
Step 5: Substitute the value of ∫₁⁰ (𝓍³ ― 2𝓍) d𝓍 from Step 3 into the equation from Step 4. This gives ∫₁⁰ (2𝓍 ― 𝓍³) d𝓍 = -3/4.
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:
1mConcetti chiave
Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.
Properties of Integrals
The properties of integrals, such as linearity and the additive property, allow us to manipulate and evaluate integrals more easily. For instance, the integral of a sum can be expressed as the sum of the integrals, and constants can be factored out. Understanding these properties is essential for simplifying complex integrals and relating them to known values.
Video consigliato:
Percorso guidato
06:21
Properties of Functions
Definite Integrals
A definite integral represents the signed area under a curve between two limits. It is calculated using the Fundamental Theorem of Calculus, which connects differentiation and integration. Knowing how to evaluate definite integrals is crucial for solving problems that involve finding the total accumulation of quantities over an interval.
Video consigliato:
Percorso guidato
05:43Definition of the Definite Integral
Substitution Method
The substitution method is a technique used to simplify the evaluation of integrals by changing variables. This method involves substituting a part of the integral with a new variable, which can make the integral easier to solve. Mastery of this technique is important for tackling integrals that are not straightforward or that involve complex expressions.
Video consigliato:
07:33Euler's Method
Pratica correlata
Domanda del libro di testo
Suppose ƒ is an even function and ∫⁸₋₈ ƒ(𝓍) d𝓍 = 18
(b) Evaluate ∫₋₈⁸ 𝓍ƒ(𝓍) d𝓍 .
58
views
Domanda del libro di testo
{Use of Tech} Functions defined by integrals Consider the function g, which is given in terms of a definite integral with a variable upper limit.
b) Calculate g'(𝓍)
g(𝓍) = ∫₀ˣ sin² t dt
65
views
Domanda del libro di testo
Area functions for linear functions Consider the following functions ƒ and real numbers a (see figure).
(b) Verify that A'(𝓍) = ƒ(𝓍).
ƒ(t) = 3t + 1 , a = 2
71
views
Domanda del libro di testo
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
(b) A left Riemann sum always overestimates the area of a region bounded by a positive increasing function and the x-axis on an interval [a,b].
81
views
Domanda del libro di testo
{Use of Tech} Midpoint Riemann sums with a calculator Consider the following definite integrals.
(b) Evaluate each sum using a calculator with n = 20, 50, and 100. Use these values to estimate the value of the integral.
∫₀⁴ (4𝓍― 𝓍²) d𝓍
86
views