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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![The image contains handwritten mathematical expressions related to integral calculus. The task is to use given properties to evaluate a specific integral.
### Problem Statement
**Task:** Use the properties to evaluate the integral.
Given:
1. \(\int_{2}^{6} x^3 \, dx = 320\)
2. \(\int_{2}^{6} x \, dx = 16\)
3. \(\int_{2}^{6} dx = 4\)
Evaluate:
\[
\int_{2}^{6} (21 - 5x - x^3) \, dx
\]
**Approach:**
To solve the given integral, you can use the linearity of integration, i.e., \(\int (f(x) + g(x) + h(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx + \int h(x) \, dx\).
1. Separate the given integral:
\[
\int_{2}^{6} 21 \, dx - 5 \int_{2}^{6} x \, dx - \int_{2}^{6} x^3 \, dx
\]
2. Use the given values to compute:
- The first integral: \(21 \times \int_{2}^{6} dx = 21 \times 4 = 84\)
- The second integral: \(-5 \times \int_{2}^{6} x \, dx = -5 \times 16 = -80\)
- The third integral: \(-\int_{2}^{6} x^3 \, dx = -320\)
3. Combine the results:
\[
84 - 80 - 320 = -316
\]
Therefore, the evaluated integral is \(-316\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb37dcdcd-0ea6-4523-b03b-c2bd2641954c%2F4a9cf2c3-401e-464b-9f48-8598f94a70e6%2Fp43xwvq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image contains handwritten mathematical expressions related to integral calculus. The task is to use given properties to evaluate a specific integral.
### Problem Statement
**Task:** Use the properties to evaluate the integral.
Given:
1. \(\int_{2}^{6} x^3 \, dx = 320\)
2. \(\int_{2}^{6} x \, dx = 16\)
3. \(\int_{2}^{6} dx = 4\)
Evaluate:
\[
\int_{2}^{6} (21 - 5x - x^3) \, dx
\]
**Approach:**
To solve the given integral, you can use the linearity of integration, i.e., \(\int (f(x) + g(x) + h(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx + \int h(x) \, dx\).
1. Separate the given integral:
\[
\int_{2}^{6} 21 \, dx - 5 \int_{2}^{6} x \, dx - \int_{2}^{6} x^3 \, dx
\]
2. Use the given values to compute:
- The first integral: \(21 \times \int_{2}^{6} dx = 21 \times 4 = 84\)
- The second integral: \(-5 \times \int_{2}^{6} x \, dx = -5 \times 16 = -80\)
- The third integral: \(-\int_{2}^{6} x^3 \, dx = -320\)
3. Combine the results:
\[
84 - 80 - 320 = -316
\]
Therefore, the evaluated integral is \(-316\).
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