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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
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Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.R.43
Capitolo 5, Problema 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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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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Definition of the Definite Integral

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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Integration by Parts for Definite Integrals

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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Fundamental Theorem of Calculus Part 1
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