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.57

Chapter 5, Problem 5.R.57

Evaluating integrals Evaluate the following integrals.

∫₀² (2𝓍 + 1)³ d𝓍

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Step 1: Recognize that the integral ∫₀² (2𝓍 + 1)³ d𝓍 involves a polynomial raised to a power. To simplify, consider using substitution. Let u = 2𝓍 + 1, which transforms the integral into a simpler form.

Step 2: Compute the derivative of u with respect to 𝓍: du/d𝓍 = 2. Rearrange to express du in terms of d𝓍: du = 2 d𝓍.

Step 3: Adjust the limits of integration to match the substitution. When 𝓍 = 0, u = 2(0) + 1 = 1. When 𝓍 = 2, u = 2(2) + 1 = 5. The integral now becomes ∫₁⁵ u³ (1/2) du, where the factor 1/2 comes from the substitution for d𝓍.

Step 4: Factor out the constant 1/2 from the integral: (1/2) ∫₁⁵ u³ du. Now, apply the power rule for integration: ∫ uⁿ du = (uⁿ⁺¹)/(n+1) + C, where n is the exponent.

Step 5: Use the power rule to integrate u³: (1/2) [(u⁴)/4] evaluated from u = 1 to u = 5. Substitute the limits into the result to complete the evaluation.

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Key Concepts

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

Definite Integrals

A definite integral represents the signed area under a curve defined by a function over a specific interval. In this case, the integral ∫₀² (2𝓍 + 1)³ d𝓍 calculates the area between the curve (2𝓍 + 1)³ and the x-axis from x = 0 to x = 2. The limits of integration (0 and 2) indicate the bounds of the area being evaluated.

Integration Techniques

To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or polynomial expansion. For the integral ∫₀² (2𝓍 + 1)³ d𝓍, expanding the integrand (2𝓍 + 1)³ into a polynomial form can simplify the integration process, making it easier to compute the definite integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links differentiation and integration, stating that if F is an antiderivative of f on an interval [a, b], then ∫ₐᵇ f(x) dx = F(b) - F(a). This theorem allows us to evaluate definite integrals by finding an antiderivative of the integrand, which is essential for solving the given integral problem.

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Related Practice

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(g) ∫ ƒ' (g(𝓍))g' (𝓍) d(𝓍) = ƒ(g(𝓍)) + C .

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(f) ∫ₐᵇ (2 ƒ(𝓍) ―3g (𝓍)) d𝓍 = 2 ∫ₐᵇ ƒ(𝓍) d𝓍 + 3 ∫₆ᵃ g(𝓍) d𝓍 .

Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume ƒ and ƒ' are continuous functions for all real numbers.

(b) Given an area function A(𝓍) = ∫ₐˣ ƒ(t) dt and an antiderivative F of ƒ, it follows that A'(𝓍) = F(𝓍) .

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Textbook Question

Evaluating integrals Evaluate the following integrals.

∫ (cos 7ω) /(16 + sin² 7ω) dω

Evaluating integrals Evaluate the following integrals.∫₀² (2𝓍 + 1)³ d𝓍

Verified video answer for a similar problem:

Key Concepts

Definite Integrals

Integration Techniques

Fundamental Theorem of Calculus

Evaluating integrals Evaluate the following integrals.

∫₀² (2𝓍 + 1)³ d𝓍