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: Calculating Area Under a Curve**
Find the area between the graph of \( y = x^3 - 6x + 8 \) and the x-axis from \( x = 1 \) to \( x = 3 \).
**Solution Approach**
To determine the area between the curve and the x-axis over the specified interval, you need to calculate the definite integral of the function \( y = x^3 - 6x + 8 \) from \( x = 1 \) to \( x = 3 \).
1. **Set Up the Integral:**
\[
\int_{1}^{3} (x^3 - 6x + 8) \, dx
\]
2. **Compute the Integral:**
- The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \).
- The antiderivative of \( -6x \) is \( -3x^2 \).
- The antiderivative of \( 8 \) is \( 8x \).
3. **Evaluate the Definite Integral:**
\[
\left[ \frac{x^4}{4} - 3x^2 + 8x \right]_{1}^{3}
\]
- Substitute \( x = 3 \) and \( x = 1 \) into the antiderivative and subtract to find the accumulated area between the curve and the x-axis over the given interval.
This procedure provides the required area under the curve from \( x = 1 \) to \( x = 3 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2442601b-b836-49c0-a398-5520e20f2138%2F288b7799-d435-4d28-bd94-cf4227b9a450%2Frotvbi_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Calculating Area Under a Curve**
Find the area between the graph of \( y = x^3 - 6x + 8 \) and the x-axis from \( x = 1 \) to \( x = 3 \).
**Solution Approach**
To determine the area between the curve and the x-axis over the specified interval, you need to calculate the definite integral of the function \( y = x^3 - 6x + 8 \) from \( x = 1 \) to \( x = 3 \).
1. **Set Up the Integral:**
\[
\int_{1}^{3} (x^3 - 6x + 8) \, dx
\]
2. **Compute the Integral:**
- The antiderivative of \( x^3 \) is \( \frac{x^4}{4} \).
- The antiderivative of \( -6x \) is \( -3x^2 \).
- The antiderivative of \( 8 \) is \( 8x \).
3. **Evaluate the Definite Integral:**
\[
\left[ \frac{x^4}{4} - 3x^2 + 8x \right]_{1}^{3}
\]
- Substitute \( x = 3 \) and \( x = 1 \) into the antiderivative and subtract to find the accumulated area between the curve and the x-axis over the given interval.
This procedure provides the required area under the curve from \( x = 1 \) to \( x = 3 \).
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