Textbook Question
Matching functions with area functions Match the functions ƒ, whose graphs are given in a― d, with the area functions A (𝓍) = ∫₀ˣ ƒ(t) dt, whose graphs are given in A–D.
Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 5, Problem 5.5.16d
Use Table 5.6 to evaluate the following definite integrals.
(d) ∫₀^π/¹⁶ sec ² 4𝓍 d𝓍
Verified step by step guidance1
Step 1: Recognize the integral ∫₀^(π/16) sec²(4𝓍) d𝓍 as a standard form integral. The integral of sec²(u) with respect to u is tan(u).
Step 2: Identify the substitution needed for the integral. Let u = 4𝓍, which implies that du = 4 d𝓍. Rewrite the integral in terms of u.
Step 3: Adjust the limits of integration based on the substitution. When 𝓍 = 0, u = 4(0) = 0. When 𝓍 = π/16, u = 4(π/16) = π/4.
Step 4: Rewrite the integral using the substitution: ∫₀^(π/16) sec²(4𝓍) d𝓍 becomes (1/4) ∫₀^(π/4) sec²(u) du, where the factor 1/4 comes from the substitution du = 4 d𝓍.
Step 5: Evaluate the integral ∫₀^(π/4) sec²(u) du using the antiderivative of sec²(u), which is tan(u). Substitute the limits of integration into tan(u) and multiply by the factor 1/4 to complete the solution.
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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. It is denoted as ∫_a^b f(x) dx, where 'a' and 'b' are the lower and upper limits, respectively. The result of a definite integral is a numerical value that quantifies the accumulation of the function's values over the interval [a, b]. Understanding how to evaluate definite integrals is crucial for solving problems in calculus.
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Integration Techniques
Integration techniques are methods used to find the integral of a function, especially when it cannot be integrated using basic formulas. Common techniques include substitution, integration by parts, and using trigonometric identities. In this context, recognizing the appropriate technique to apply is essential for evaluating the integral effectively, particularly when dealing with functions like sec²(4x).
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Trigonometric Functions
Trigonometric functions, such as secant (sec), are fundamental in calculus and relate angles to ratios of sides in right triangles. The secant function is defined as the reciprocal of the cosine function, sec(x) = 1/cos(x). Understanding the properties and derivatives of trigonometric functions is vital for evaluating integrals involving these functions, as they often appear in various calculus problems.
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Related Practice
Textbook Question
Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.
{Use of Tech} ƒ(𝓍) = √x on [1,3] ; n = 4
(d) Calculate the midpoint Riemann sum.
Textbook Question
Left and right Riemann sums Complete the following steps for the given function, interval, and value of n.
ƒ(𝓍) = x² ─ 1 on [2,4]; n = 4
(d) Calculate the left and right Riemann sums.
Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
(d) If ∫ₐᵇ ƒ(𝓍) d𝓍 = ∫ₐᵇ ƒ(𝓍) d𝓍, then ƒ is a constant function.
Textbook Question
Properties of integrals Suppose ∫₀³ƒ(𝓍) d𝓍 = 2 , ∫₃⁶ƒ(𝓍) d𝓍 = ―5 , and ∫₃⁶g(𝓍) d𝓍 = 1. Evaluate the following integrals.
(a) ∫₀³ 5ƒ(𝓍) d𝓍
Textbook Question
Midpoint Riemann sums Complete the following steps for the given function, interval, and value of n.
ƒ(𝓍) = 1/x on [1,6] ; n = 5
(d) Calculate the midpoint Riemann sum.