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2. Use a geometric formula to determine the area of the region given by f: V25 – x5dx. -5

2. Use a geometric formula to determine the area of the region given by f: V25 – x5dx. -5

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Problem 2:** Use a geometric formula to determine the area of the region given by the integral: \[ \int_{-5}^{5} \sqrt{25 - x^2} \, dx. \] This problem involves finding the area enclosed under the curve defined by the function \(y = \sqrt{25 - x^2}\) from \(x = -5\) to \(x = 5\). The expression \(\sqrt{25 - x^2}\) represents the equation of a semicircle with radius 5 centered on the x-axis. Therefore, the area can be calculated using the formula for the area of a circle (\(\pi r^2\)), considering only the upper half: - **Radius**: 5 - **Area of the full circle**: \(\pi \times 5^2 = 25\pi\) - **Area of the semicircle**: \(\frac{1}{2} \times 25\pi = \frac{25\pi}{2}\) Thus, the area of the region described by the integral is \(\frac{25\pi}{2}\).
Transcribed Image Text:**Problem 2:** Use a geometric formula to determine the area of the region given by the integral: \[ \int_{-5}^{5} \sqrt{25 - x^2} \, dx. \] This problem involves finding the area enclosed under the curve defined by the function \(y = \sqrt{25 - x^2}\) from \(x = -5\) to \(x = 5\). The expression \(\sqrt{25 - x^2}\) represents the equation of a semicircle with radius 5 centered on the x-axis. Therefore, the area can be calculated using the formula for the area of a circle (\(\pi r^2\)), considering only the upper half: - **Radius**: 5 - **Area of the full circle**: \(\pi \times 5^2 = 25\pi\) - **Area of the semicircle**: \(\frac{1}{2} \times 25\pi = \frac{25\pi}{2}\) Thus, the area of the region described by the integral is \(\frac{25\pi}{2}\).
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