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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.R.43

Evaluating integrals Evaluate the following integrals.


∫₀¹ √𝓍 (√𝓍 + 1) d𝓍

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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.
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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.
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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

Evaluating integrals Evaluate the following integrals.


∫₀^²π cos² 𝓍/6 d𝓍

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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.

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Textbook Question

Evaluate the following derivatives.


d/d𝓍 ∫₃ᵉˣ cos t² dt

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Textbook Question

Function defined by an integral Let ƒ(𝓍) = ∫₀ˣ (t ― 1)¹⁵ (t―2)⁹ dt .

(c) For what values of 𝓍 does ƒ have local minima? Local maxima?

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Textbook Question

Geometry of integrals Without evaluating the integrals, explain why the following statement is true for positive integers n:

∫₀¹ 𝓍ⁿd𝓍 + ∫₀¹ ⁿ√(𝓍d𝓍) = 1

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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) = 2 sin 𝓍/4 on [0, 2π]

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