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Ch. 5 - Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 5 - IntegrationProblema 5.5.40
Capitolo 5, Problema 5.5.40

Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.                                                                                  
                                                                                                                                                                    
 ∫ (sin⁵ 𝓍 + 3 sin³ 𝓍― sin 𝓍) cos 𝓍 d𝓍

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Step 1: Recognize that the integral involves powers of sin(𝓍) multiplied by cos(𝓍). This suggests using a substitution method where u = sin(𝓍).
Step 2: Compute the derivative of u with respect to 𝓍: du/d𝓍 = cos(𝓍), which implies that du = cos(𝓍) d𝓍. Substitute u = sin(𝓍) and du = cos(𝓍) d𝓍 into the integral.
Step 3: Rewrite the integral in terms of u: ∫ (u⁵ + 3u³ - u) du. This simplifies the integral into a polynomial form.
Step 4: Apply the power rule for integration to each term of the polynomial: ∫ u⁵ du = (u⁶)/6, ∫ 3u³ du = (3u⁴)/4, and ∫ -u du = -(u²)/2.
Step 5: Combine the results to express the indefinite integral in terms of u, then substitute back u = sin(𝓍) to return to the original variable. Finally, check your work by differentiating the result to ensure it matches the original integrand.

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Indefinite Integrals

Indefinite integrals represent a family of functions whose derivative gives the integrand. They are expressed without limits and include a constant of integration, typically denoted as 'C'. The process of finding an indefinite integral is often referred to as antiderivation, and it is essential for solving problems in calculus involving area under curves and accumulation of quantities.
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Introduction to Indefinite Integrals

Change of Variables

Change of variables, or substitution, is a technique used in integration to simplify the integrand. By substituting a new variable for a function of the original variable, the integral can often be transformed into a more manageable form. This method is particularly useful when dealing with complex functions or when the integrand can be expressed in terms of a simpler function.
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Changing Geometries

Differentiation Check

Checking work by differentiation involves taking the derivative of the result obtained from an indefinite integral to verify its correctness. This process ensures that the original integrand is recovered, confirming that the integration was performed accurately. It serves as a crucial step in validating the solution and reinforcing the relationship between differentiation and integration.
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Derivatives of integrals Simplify the following expressions.


d/dt ∫₀ᵗ d𝓍/(1 + 𝓍²) + ∫₁¹/ᵗ dx/(1 + 𝓍²)

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Definite integrals from graphs The figure shows the areas of regions bounded by the graph of ƒ and the 𝓍-axis. Evaluate the following integrals.



∫₀ᵃ ƒ(𝓍) d𝓍

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Average value of the derivative Suppose ƒ ' is a continuous function for all real numbers. Show that the average value of the derivative on an interval [a, b] is ƒ⁻' = (ƒ(b) ―ƒ(a))/ (b―a) . Interpret this result in terms of secant lines.

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General results Evaluate the following integrals in which the function ƒ is unspecified. Note that ƒ⁽ᵖ⁾ is the pth derivative of ƒ and ƒᵖ is the pth power of ƒ. Assume ƒ and its derivatives are continuous for all real numbers. 

∫ (5 ƒ³ (𝓍) + 7ƒ² (𝓍) + ƒ (𝓍 )) ƒ'(𝓍) d𝓍

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Definite integrals Use a change of variables or Table 5.6 to evaluate the following definite integrals.                                                                                                                         

                                                                                                                                                                              

 ∫₋₁¹ (𝓍―1) (𝓍²―2𝓍)⁷ d𝓍

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Use the given substitution to evaluate the following indefinite integrals. Check your answer by differentiating.                                                                                              

                                                                                                                                                                                        

 ∫ (6𝓍 + 1) √(3𝓍² + 𝓍) d𝓍 , u = 3𝓍² + 𝓍

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