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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 5.4.46a

Planetary orbits The planets orbit the Sun in elliptical orbits with the Sun at one focus (see Section 12.4 for more on ellipses). The equation of an ellipse whose dimensions are 2a in the 𝓍-direction and 2b in the y-direction is (𝓍²/a²) + (y² /b²) = 1.
(a) Let d² denote the square of the distance from a planet to the center of the ellipse at (0, 0). Integrate over the interval [ ―a, a] to show that the average value of d² is (a² + 2b²) /3 .
Diagram of an elliptical orbit with the Sun at one focus, labeled axes, and the Earth positioned along the orbit.

Verified step by step guidance
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Step 1: Begin by recalling the equation of the ellipse: (𝓍²/a²) + (y²/b²) = 1. This represents the elliptical orbit of the planet with semi-major axis 'a' and semi-minor axis 'b'.
Step 2: The square of the distance from the planet to the center of the ellipse is given by d² = 𝓍² + y². Substitute y² from the ellipse equation into this expression: y² = b²(1 - 𝓍²/a²). Thus, d² = 𝓍² + b²(1 - 𝓍²/a²).
Step 3: Simplify the expression for d²: d² = 𝓍² + b² - (b²𝓍²/a²). Combine terms to get d² = (1 - b²/a²)𝓍² + b².
Step 4: To find the average value of d² over the interval [−a, a], use the formula for the average value of a function: (1/(a - (-a))) ∫[−a, a] d² d𝓍. This simplifies to (1/(2a)) ∫[−a, a] [(1 - b²/a²)𝓍² + b²] d𝓍.
Step 5: Break the integral into two parts: (1/(2a)) [∫[−a, a] (1 - b²/a²)𝓍² d𝓍 + ∫[−a, a] b² d𝓍]. Evaluate each integral separately, noting that ∫[−a, a] 𝓍² d𝓍 = (2/3)a³ and ∫[−a, a] d𝓍 = 2a. Combine results to show that the average value of d² is (a² + 2b²)/3.

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Key Concepts

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

Elliptical Orbits

Elliptical orbits describe the paths that celestial bodies, such as planets, take around a focal point, like the Sun. An ellipse is defined by its semi-major axis (a) and semi-minor axis (b), with the Sun located at one of the foci. Understanding the properties of ellipses is crucial for analyzing planetary motion and calculating distances within these orbits.

Average Value of a Function

The average value of a function over a given interval is calculated by integrating the function over that interval and then dividing by the length of the interval. In this context, the average value of the square of the distance from a planet to the center of the ellipse is derived through integration, which provides insights into the planet's position relative to the Sun over its orbital path.
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Integration

Integration is a fundamental concept in calculus that involves finding the accumulated area under a curve represented by a function. In this problem, integration is used to compute the average value of the squared distance from the center of the ellipse, which requires setting up the appropriate integral and evaluating it over the specified interval. Mastery of integration techniques is essential for solving problems related to areas, volumes, and averages in calculus.
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∫₀⁴ (4𝓍― 𝓍²) d𝓍

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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.



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a) Divide the interval [0,8] into n = 2 subintervals, [0,4] and [4,8]. On each subinterval, assume the object moves at a constant velocity equal to the value of v evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [0,8] (see part (a) of the figure)                                                                                                             


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Working with area functions Consider the function ƒ and the points a, b, and c.

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Area functions The graph of ƒ is shown in the figure. Let A(x) = ∫₋₂ˣ ƒ(t) dt and F(x) = ∫₄ˣ ƒ(t) dt be two area functions for ƒ. Evaluate the following area functions.

(a) A (―2)

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