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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Concept explainers
Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question
![### Integral Problem
#### Problem 3
Evaluate the following integral:
\[ \int (x - \sin(2x))^2 \, dx \]
Explanation: This integral involves a quadratic expression where the argument of the sine function is twice the variable \(x\). The challenge here is to simplify the integrand and then integrate term by term.
#### Solution Outline
1. **Expand the Expression**:
First, expand the square term inside the integral.
\[ (x - \sin(2x))^2 = x^2 - 2x\sin(2x) + \sin^2(2x) \]
2. **Integrate Each Term Separately**:
- Integrate \( x^2 \)
- Integrate \( -2x\sin(2x) \)
- Integrate \( \sin^2(2x) \)
Here are the steps to integrate each term, broken down:
1. **Integral of \( x^2 \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
2. **Integral of \( -2x\sin(2x) \)**:
This part can be solved using integration by parts.
Let:
\[ u = x \quad \text{and} \quad dv = -2\sin(2x) \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = \int -2\sin(2x) \, dx = \int -2 \cdot \frac{-1}{2} \cos(2x) = \cos(2x) \]
Using integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Therefore:
\[ \int -2x\sin(2x) \, dx = x\cos(2x) - \int \cos(2x) \, dx \]
\[ = x\cos(2x) - \frac{1}{2}\sin(2x) \]
3. **Integral of \( \sin^2(2x) \)**:
Use the trigonometric identity to simplify:
\[ \sin^2(2x) = \frac{1 -](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F502330eb-9fa1-41eb-8716-374edceb5782%2F23da6f6b-1619-4142-b25b-e3f36aefbd1b%2Flut2nj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Integral Problem
#### Problem 3
Evaluate the following integral:
\[ \int (x - \sin(2x))^2 \, dx \]
Explanation: This integral involves a quadratic expression where the argument of the sine function is twice the variable \(x\). The challenge here is to simplify the integrand and then integrate term by term.
#### Solution Outline
1. **Expand the Expression**:
First, expand the square term inside the integral.
\[ (x - \sin(2x))^2 = x^2 - 2x\sin(2x) + \sin^2(2x) \]
2. **Integrate Each Term Separately**:
- Integrate \( x^2 \)
- Integrate \( -2x\sin(2x) \)
- Integrate \( \sin^2(2x) \)
Here are the steps to integrate each term, broken down:
1. **Integral of \( x^2 \)**:
\[ \int x^2 \, dx = \frac{x^3}{3} \]
2. **Integral of \( -2x\sin(2x) \)**:
This part can be solved using integration by parts.
Let:
\[ u = x \quad \text{and} \quad dv = -2\sin(2x) \, dx \]
Then:
\[ du = dx \quad \text{and} \quad v = \int -2\sin(2x) \, dx = \int -2 \cdot \frac{-1}{2} \cos(2x) = \cos(2x) \]
Using integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
Therefore:
\[ \int -2x\sin(2x) \, dx = x\cos(2x) - \int \cos(2x) \, dx \]
\[ = x\cos(2x) - \frac{1}{2}\sin(2x) \]
3. **Integral of \( \sin^2(2x) \)**:
Use the trigonometric identity to simplify:
\[ \sin^2(2x) = \frac{1 -
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