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Indefinite integrals Use a change of variables or Table 5.6 to evaluate the following indefinite integrals. Check your work by differentiating.
∫ sin 𝓍 sec⁸ 𝓍 d𝓍
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Ch. 5 - Integration
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 5.4.15
Symmetry in integrals Use symmetry to evaluate the following integrals.
∫²₋₂ (x² + x³) dx
Guida verificata passo dopo passo1
Step 1: Recognize the integral bounds are symmetric about the origin, as they range from -2 to 2. This suggests that symmetry properties of the function can be used to simplify the evaluation.
Step 2: Analyze the integrand, x² + x³. Split it into two terms: x² and x³. Determine the symmetry of each term. x² is an even function because f(-x) = (-x)² = x², and x³ is an odd function because f(-x) = (-x)³ = -x³.
Step 3: Use the property of integrals for symmetric bounds. For even functions, ∫₋ₐᵃ f(x) dx = 2∫₀ᵃ f(x) dx, and for odd functions, ∫₋ₐᵃ f(x) dx = 0. Apply these properties to the terms x² and x³.
Step 4: Since x² is even, its integral over [-2, 2] simplifies to 2∫₀² x² dx. Since x³ is odd, its integral over [-2, 2] is 0. Combine these results to simplify the original integral.
Step 5: Write the simplified integral as ∫²₋₂ (x² + x³) dx = 2∫₀² x² dx + 0. Evaluate ∫₀² x² dx using standard integration techniques to find the final result.
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Symmetry in Functions
Symmetry in functions refers to the property where a function exhibits even or odd characteristics. An even function, such as f(x) = f(-x), is symmetric about the y-axis, while an odd function, f(x) = -f(-x), is symmetric about the origin. Recognizing these properties can simplify the evaluation of integrals, particularly over symmetric intervals.
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06:21
Properties of Functions
Definite Integrals
A definite integral calculates the area under a curve defined by a function over a specific interval [a, b]. The result is a numerical value representing this area. Understanding how to set up and evaluate definite integrals is crucial for applying symmetry properties effectively, as it allows for the simplification of calculations when the function's symmetry is identified.
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05:43Definition of the Definite Integral
Properties of Integrals
The properties of integrals include linearity, which allows the integral of a sum to be the sum of the integrals, and the ability to evaluate integrals over symmetric intervals. For example, if a function is odd and integrated over a symmetric interval around zero, the integral evaluates to zero. These properties are essential for efficiently solving integrals that exhibit symmetry.
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06:21Properties of Functions
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