11) Define A(k) = x dx Using area, create a nice formula for A(k). Explain your area reasoning. Using area, explain why f" sin(0) de = 0 Using area, explain why cos(0) de = 0
11) Define A(k) = x dx Using area, create a nice formula for A(k). Explain your area reasoning. Using area, explain why f" sin(0) de = 0 Using area, explain why cos(0) de = 0
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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![### Transcription and Explanation
#### Problem 11
**Define**
\[ A(k) = \int_{0}^{k} x \, dx \]
Using area, create a nice formula for \( A(k) \). Explain your area reasoning.
**Explanation:**
To find the area under the curve \( y = x \) from \( x = 0 \) to \( x = k \), we recognize this as the area of a right triangle with a base of \( k \) and a height of \( k \).
The area \( A \) of a triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values, we get:
\[ A(k) = \frac{1}{2} \times k \times k = \frac{k^2}{2} \]
Thus, \( A(k) = \frac{k^2}{2} \).
---
**Using area, explain why**
\[ \int_{-\pi}^{\pi} \sin(\theta) \, d\theta = 0 \]
**Explanation:**
The function \( \sin(\theta) \) is symmetric about the origin, having the same area above the x-axis as it does below within the interval \([-π, π]\). The positive area from \([-π, 0]\) cancels out with the negative area from \([0, π]\). Therefore, the total area is zero.
---
**Using area, explain why**
\[ \int_{0}^{\pi} \cos(\theta) \, d\theta = 0 \]
**Explanation:**
The function \( \cos(\theta) \) from \( 0 \) to \( \pi \) has equal positive and negative areas. \( \cos(\theta) \) is positive over \([0, \frac{\pi}{2}]\) and negative over \([\frac{\pi}{2}, \pi]\). These areas are equal in magnitude, thus cancel out, making the integral zero.
---
**Using area, explain why**
\[ \int_{0}^{2} (t+5)(t-3) \, dt < 0 \]
**Explanation:**
First, expand the expression:
\[ (t+5)(t-3) = t^2 - 3t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb5f1630e-c465-44f2-a868-b429d99d07e1%2F7a71a9b8-c392-4c34-b60a-f2261fc9b358%2F7um8c7_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription and Explanation
#### Problem 11
**Define**
\[ A(k) = \int_{0}^{k} x \, dx \]
Using area, create a nice formula for \( A(k) \). Explain your area reasoning.
**Explanation:**
To find the area under the curve \( y = x \) from \( x = 0 \) to \( x = k \), we recognize this as the area of a right triangle with a base of \( k \) and a height of \( k \).
The area \( A \) of a triangle is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the values, we get:
\[ A(k) = \frac{1}{2} \times k \times k = \frac{k^2}{2} \]
Thus, \( A(k) = \frac{k^2}{2} \).
---
**Using area, explain why**
\[ \int_{-\pi}^{\pi} \sin(\theta) \, d\theta = 0 \]
**Explanation:**
The function \( \sin(\theta) \) is symmetric about the origin, having the same area above the x-axis as it does below within the interval \([-π, π]\). The positive area from \([-π, 0]\) cancels out with the negative area from \([0, π]\). Therefore, the total area is zero.
---
**Using area, explain why**
\[ \int_{0}^{\pi} \cos(\theta) \, d\theta = 0 \]
**Explanation:**
The function \( \cos(\theta) \) from \( 0 \) to \( \pi \) has equal positive and negative areas. \( \cos(\theta) \) is positive over \([0, \frac{\pi}{2}]\) and negative over \([\frac{\pi}{2}, \pi]\). These areas are equal in magnitude, thus cancel out, making the integral zero.
---
**Using area, explain why**
\[ \int_{0}^{2} (t+5)(t-3) \, dt < 0 \]
**Explanation:**
First, expand the expression:
\[ (t+5)(t-3) = t^2 - 3t
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