Ch. 5 - Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 5 - Integration

Problem 5.R.78

Chapter 5, Problem 5.R.78

Evaluating integrals Evaluate the following integrals.

∫ 𝓍⁷ √(𝓍⁴ + 1d𝓍)

Verified step by step guidance

Step 1: Recognize that the integral involves a composite function. The term √(𝓍⁴ + 1) suggests a substitution method might simplify the integral. Let u = 𝓍⁴ + 1.

Step 2: Compute the derivative of u with respect to 𝓍. Since u = 𝓍⁴ + 1, we find du/d𝓍 = 4𝓍³. Rearrange to express du in terms of d𝓍: du = 4𝓍³ d𝓍.

Step 3: Rewrite the integral in terms of u. Substitute u = 𝓍⁴ + 1 and du = 4𝓍³ d𝓍 into the original integral. The integral becomes (1/4) ∫ 𝓍⁴ √u du.

Step 4: Simplify further. Notice that 𝓍⁴ can be expressed in terms of u using the substitution u = 𝓍⁴ + 1. Therefore, 𝓍⁴ = u - 1. Replace 𝓍⁴ in the integral to get (1/4) ∫ (u - 1) √u du.

Step 5: Break the integral into two simpler parts. Expand (u - 1) √u into u^(3/2) - u^(1/2) and integrate each term separately. Use the power rule for integration: ∫ u^n du = (u^(n+1))/(n+1) + C, where n ≠ -1.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Integration

Integration is a fundamental concept in calculus that involves finding the integral of a function, which represents the area under the curve of that function on a given interval. It can be thought of as the reverse process of differentiation. There are various techniques for integration, including substitution, integration by parts, and numerical methods, each suited for different types of integrals.

Substitution Method

The substitution method is a technique used in integration to simplify the process by changing the variable of integration. This involves substituting a part of the integral with a new variable, which can make the integral easier to evaluate. For example, if the integral contains a composite function, substituting the inner function can lead to a simpler integral that is easier to solve.

Definite vs. Indefinite Integrals

Integrals can be classified as definite or indefinite. An indefinite integral represents a family of functions and includes a constant of integration, while a definite integral calculates the net area under the curve between two specific limits. Understanding the difference is crucial for correctly interpreting the results of integration and applying the appropriate techniques.

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Related Practice

Textbook Question

Area of regions Compute the area of the region bounded by the graph of ƒ and the 𝓍-axis on the given interval. You may find it useful to sketch the region.

ƒ(𝓍) = 16―𝓍² on [―4, 4]

105

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Textbook Question

Find the average value of ƒ(𝓍) = e²ˣ on [0, ln 2] .

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Textbook Question

Properties of integrals Suppose ∫₁⁴ ƒ(𝓍) d𝓍 = 6 , ∫₁⁴ g(𝓍) d𝓍 = 4 and ∫₃⁴ ƒ(𝓍) d𝓍 = 2 . Evaluate the following integrals or state that there is not enough information.

∫₁³ ƒ(𝓍)/g(𝓍) d𝓍

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Textbook Question

Area functions and the Fundamental Theorem Consider the function

ƒ(t) = { t if ―2 ≤ t < 0

t²/2 if 0 ≤ t ≤ 2

and its graph shown below. Let F(𝓍) = ∫₋₁ˣ ƒ(t) dt and G(𝓍) = ∫₋₂ˣ ƒ(t) dt.

(e) Evaluate F ''(―1) and F ''(1). Interpret these values.

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Textbook Question

Area by geometry Use geometry to evaluate the following definite integrals, where the graph of ƒ is given in the figure.

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