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.43

Chapter 5, Problem 5.R.43

Evaluating integrals Evaluate the following integrals.

∫₀¹ √𝓍 (√𝓍 + 1) d𝓍

Verified step by step guidance

Step 1: Begin by expanding the integrand. The given integral is ∫₀¹ √𝓍 (√𝓍 + 1) d𝓍. Distribute √𝓍 across the terms inside the parentheses to rewrite the integrand as ∫₀¹ (𝓍 + √𝓍) d𝓍.

Step 2: Split the integral into two separate integrals for easier computation: ∫₀¹ (𝓍 + √𝓍) d𝓍 = ∫₀¹ 𝓍 d𝓍 + ∫₀¹ √𝓍 d𝓍.

Step 3: Evaluate the first integral, ∫₀¹ 𝓍 d𝓍. Use the power rule for integration, which states ∫𝓍ⁿ d𝓍 = (𝓍ⁿ⁺¹)/(n+1) + C, where n ≠ -1. Here, n = 1, so ∫₀¹ 𝓍 d𝓍 = [𝓍²/2]₀¹.

Step 4: Evaluate the second integral, ∫₀¹ √𝓍 d𝓍. Rewrite √𝓍 as 𝓍^(1/2) and apply the power rule for integration. For n = 1/2, ∫𝓍^(1/2) d𝓍 = (𝓍^(3/2))/(3/2) + C. Simplify to (2/3)𝓍^(3/2). Evaluate this expression from 0 to 1: [(2/3)𝓍^(3/2)]₀¹.

Step 5: Combine the results of the two integrals. Add the evaluated results of ∫₀¹ 𝓍 d𝓍 and ∫₀¹ √𝓍 d𝓍 to obtain the final value of the integral ∫₀¹ √𝓍 (√𝓍 + 1) d𝓍.

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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 between two specified limits. In this case, the integral ∫₀¹ √𝓍 (√𝓍 + 1) d𝓍 is evaluated from 0 to 1, which means we are calculating the area under the curve of the function √𝓍 (√𝓍 + 1) from x = 0 to x = 1.

Integration Techniques

To evaluate integrals, various techniques can be employed, such as substitution, integration by parts, or recognizing patterns. For the given integral, simplifying the integrand √𝓍 (√𝓍 + 1) may involve expanding the expression or using substitution to make the integration process more manageable.

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 is essential for evaluating definite integrals, as it allows us to find the area under the curve by calculating the difference of the antiderivative at the upper and lower limits.

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Related Practice

Textbook Question

Area versus net area Find (i) the net area and (ii) 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) = 𝓍⁴ ― 𝓍² on [―1, 1]

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Textbook Question

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

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.

(a) A(𝓍) = ∫ₐˣ ƒ(t) dt and ƒ(t) = 2t―3 , then A is a quadratic function.

Textbook Question

Evaluating integrals Evaluate the following integrals.

∫₀¹ 𝓍 • 2ˣ²⁺¹ d𝓍

Textbook Question

Use geometry and properties of integrals to evaluate the following definite integrals.

∫₄⁰ (2𝓍 + √(16―𝓍²)) d𝓍 . (Hint: Write the integral as sum of two integrals.)

Textbook Question

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

(d) ∫₀⁷ ƒ(𝓍) d𝓍

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