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 5.5.106

Chapter 5, Problem 5.5.106

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

Verified step by step guidance

Recognize that the integral is of the form \( \int (5 f^3(x) + 7 f^2(x) + f(x)) f'(x) \, dx \), where \( f(x) \) is an unspecified function and \( f'(x) \) is its derivative.

Use the substitution method by letting \( u = f(x) \). Then, the differential \( du = f'(x) \, dx \). This transforms the integral into \( \int (5 u^3 + 7 u^2 + u) \, du \).

Rewrite the integral in terms of \( u \): \( \int (5 u^3 + 7 u^2 + u) \, du = \int 5 u^3 \, du + \int 7 u^2 \, du + \int u \, du \).

Integrate each term separately using the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} + C \). So, compute \( \int 5 u^3 \, du \), \( \int 7 u^2 \, du \), and \( \int u \, du \) accordingly.

After integrating, substitute back \( u = f(x) \) to express the result in terms of \( f(x) \). Don't forget to add the constant of integration \( + C \).

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Key Concepts

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

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. When an integral contains a function and its derivative, substituting the inner function with a new variable can transform the integral into a basic form, making it easier to evaluate.

Notation and Interpretation of Derivatives and Powers

Understanding the notation is crucial: ƒ⁽ᵖ⁾ denotes the pth derivative of ƒ, while ƒᵖ means the pth power of ƒ. Distinguishing between these helps correctly interpret the integrand and apply appropriate integration techniques.

Properties of Continuous Functions and Their Derivatives

Assuming ƒ and its derivatives are continuous ensures the validity of applying substitution and other integration rules. Continuity guarantees no abrupt changes, allowing the integral to be evaluated smoothly over the real numbers.

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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

Textbook Question

{Use of Tech} Areas of regions Find the area of the region 𝑅 bounded by the graph of ƒ and the 𝓍-axis on the given interval. Graph ƒ and show the region 𝑅.

ƒ(𝓍) = 2 ― |𝓍| on [ ― 2 , 4]

Textbook Question

Why can the constant of integration be omitted from the antiderivative when evaluating a definite integral?

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] ?

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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Textbook Question

Suppose the interval [1, 3] is partitioned into n = 4 subintervals. What is the subinterval length ∆𝓍? List the grid points x₀ , x₁ , x₂ , x₃ and x₄. Which points are used for the left, right, and midpoint Riemann sums?

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