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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 5.5.80

Variations on the substitution method Evaluate the following integrals.                                                                                                        
                                                                                                                                                                    
 ∫ y²/(y + 1)⁴ dy

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1
Step 1: Identify the substitution method. Let u = y + 1, which simplifies the denominator. Then, du = dy.
Step 2: Rewrite y in terms of u. Since u = y + 1, we have y = u - 1.
Step 3: Substitute into the integral. Replace y² with (u - 1)² and (y + 1)⁴ with u⁴, and dy with du. The integral becomes ∫ (u - 1)² / u⁴ du.
Step 4: Expand (u - 1)² to simplify the numerator. This gives (u - 1)² = u² - 2u + 1. Substitute this into the integral, resulting in ∫ (u² - 2u + 1) / u⁴ du.
Step 5: Break the integral into separate terms. Divide each term in the numerator by u⁴: ∫ (u²/u⁴ - 2u/u⁴ + 1/u⁴) du, which simplifies to ∫ (1/u² - 2/u³ + 1/u⁴) du. Now, integrate each term individually.

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Key Concepts

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

Substitution Method

The substitution method is a technique used in integration to simplify the process by changing the variable of integration. By substituting a part of the integrand with a new variable, the integral can often be transformed into a more manageable form. This method is particularly useful when dealing with composite functions or when the integrand contains a function and its derivative.
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Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation, used to integrate products of functions. It states that the integral of a product of two functions can be expressed as the product of one function and the integral of the other, minus the integral of the derivative of the first function times the second. This method is helpful when the integrand is a product that can be simplified through differentiation.
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Polynomial Long Division

Polynomial long division is a method used to divide a polynomial by another polynomial of lower degree. This technique is particularly useful when the integrand is a rational function, allowing the integrand to be expressed as a sum of a polynomial and a proper fraction. By simplifying the integrand in this way, it becomes easier to integrate each term separately, facilitating the overall integration process.
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Related Practice
Textbook Question

Area Find (i) the net area and (ii) the area of the following regions. Graph the function and indicate the region in question.


The region bounded by y = 6 cos 𝓍 and the 𝓍-axis between 𝓍 = ―π/2 and 𝓍 = π

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Textbook Question

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

v = [1 / (2t + 1)] (m/s), for 0 ≤ t ≤ 8 ; n = 4

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Textbook Question

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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Textbook Question

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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Textbook Question

Average distance on a parabola What is the average distance between the parabola y = 30𝓍 (20 ― 𝓍 ) and the 𝓍-axis on the interval [0, 20] ?

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Textbook Question

Approximating displacement The velocity of an object is given by the following functions on a specified interval. Approximate the displacement of the object on this interval by subdividing the interval into n subintervals. Use the left endpoint of each subinterval to compute the height of the rectangles.

v = 2t + 1(m/s), for 0 ≤ t ≤ 8 ; n = 2

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