Textbook Question
Evaluate the integrals in Exercises 1–14.
∫ 5 dx / √(25x² - 9), where x > 3/5
Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.4.58
Area: Find the area enclosed by the ellipse x²/a² + y²/b² = 1.
Verified step by step guidance1
Recognize that the ellipse is given by the equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), where \(a\) and \(b\) are the semi-major and semi-minor axes respectively.
Express \(y\) in terms of \(x\) by solving the ellipse equation for \(y\): \(y = \pm b \sqrt{1 - \frac{x^{2}}{a^{2}}}\).
Set up the integral for the area of the ellipse by integrating the positive half of \(y\) from \(-a\) to \(a\) and then doubling the result to account for the negative half: \(\text{Area} = 2 \int_{-a}^{a} b \sqrt{1 - \frac{x^{2}}{a^{2}}} \, dx\).
Use a substitution to simplify the integral, such as \(x = a \sin \theta\), which transforms the integral limits and the integrand into a trigonometric form.
Evaluate the integral using the substitution and then multiply by 2 and \(b\) to find the total area enclosed by the ellipse.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Equation of an Ellipse
The ellipse is defined by the equation x²/a² + y²/b² = 1, where 'a' and 'b' are the semi-major and semi-minor axes respectively. This equation describes all points (x, y) that lie on the ellipse, forming a closed curve symmetric about both axes.
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Parameterizing Equations of Circles & Ellipses
Area Formula for an Ellipse
The area enclosed by an ellipse is given by the formula A = πab, where 'a' and 'b' are the lengths of the semi-major and semi-minor axes. This formula is derived by scaling the area of a unit circle by the factors 'a' and 'b' along the x and y axes.
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Integration in Calculus for Area
Calculus can be used to find the area enclosed by curves by integrating the function that defines the boundary. For an ellipse, one can set up an integral using the ellipse equation to find the area, confirming the formula A = πab through definite integration.
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Related Practice
Textbook Question
Evaluate the integrals in Exercises 39–54.
∫ sin(θ) dθ / (cos²θ + cos θ - 2)
Textbook Question
Evaluate the integrals in Exercises 33–52.
∫ from -π/4 to π/4 of 6 tan⁴(x) dx
1
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Textbook Question
For Exercises 49–52, complete the square before using an appropriate trigonometric substitution.
∫ 1 / √(x² - 2x + 5) dx
Textbook Question
Use any method to evaluate the integrals in Exercises 55–66.
∫ 2 / (x(ln x - 2)³) dx
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Textbook Question
Evaluate the integrals in Exercises 1–22.
∫ sin⁴(2x) cos(2x) dx