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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Question
![**Problem Statement**:
Find the value without finding the antiderivative.
\[
\int_{0}^{5} \sqrt{25 - x^2} \, dx
\]
**Instructions**:
- Evaluate the integral using geometric methods.
- You can find a hint in the tutorial provided if needed.
**Visualization and Context**:
- The integral \(\int_{0}^{5} \sqrt{25 - x^2} \, dx\) represents the area under the curve of the function \(\sqrt{25 - x^2}\) from \(x = 0\) to \(x = 5\).
- This graph describes a semicircle with radius 5 centered at the origin (\(x = 0\)) and stretching to \(x = 5\).
**Tip**:
- Use the geometric property of the semicircle to find the area. Remember the formula for the area of a circle is \(A = \pi r^2\).
Click the "Tutorial" button for a step-by-step guide.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7e02a959-93d9-49eb-9331-56d698ab2563%2Fba16e35c-6efd-4255-aab8-fb13540c466a%2F1huax2_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**:
Find the value without finding the antiderivative.
\[
\int_{0}^{5} \sqrt{25 - x^2} \, dx
\]
**Instructions**:
- Evaluate the integral using geometric methods.
- You can find a hint in the tutorial provided if needed.
**Visualization and Context**:
- The integral \(\int_{0}^{5} \sqrt{25 - x^2} \, dx\) represents the area under the curve of the function \(\sqrt{25 - x^2}\) from \(x = 0\) to \(x = 5\).
- This graph describes a semicircle with radius 5 centered at the origin (\(x = 0\)) and stretching to \(x = 5\).
**Tip**:
- Use the geometric property of the semicircle to find the area. Remember the formula for the area of a circle is \(A = \pi r^2\).
Click the "Tutorial" button for a step-by-step guide.
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