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 total area between \( y = 25 - x^2 \) and the x-axis for \( 0 \leq x \leq 6 \).
**Solution:**
To find the area between the curve \( y = 25 - x^2 \) and the x-axis from \( x = 0 \) to \( x = 6 \), we need to set up and evaluate the definite integral of the function \( 25 - x^2 \) with respect to \( x \) over the given interval.
The definite integral for this problem is:
\[
\int_{0}^{6} (25 - x^2) \, dx
\]
### Step-by-Step Solution:
1. **Set up the Integral:**
\[
A = \int_{0}^{6} (25 - x^2) \, dx
\]
2. **Integrate the Function:**
\[
\int (25 - x^2) \, dx = 25x - \frac{x^3}{3}
\]
3. **Evaluate the Integral at the Bounds:**
\[
A = \left[ 25x - \frac{x^3}{3} \right]_{0}^{6}
\]
4. **Calculate the Values:**
\[
A = \left(25(6) - \frac{6^3}{3}\right) - \left(25(0) - \frac{0^3}{3}\right)
\]
\[
= \left(150 - 72\right) - (0)
\]
\[
= 78
\]
Thus, the total area is:
\[
\boxed{78}
\]
### Conclusion:
The total area between the curve \( y = 25 - x^2 \) and the x-axis for \( 0 \leq x \leq 6 \) is \( 78 \) square units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdf0e5cd0-c292-48e9-a1f1-41b5e15042f7%2F3c4bdf97-9de0-44a1-a304-f3113146b494%2F1mvnzmn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the total area between \( y = 25 - x^2 \) and the x-axis for \( 0 \leq x \leq 6 \).
**Solution:**
To find the area between the curve \( y = 25 - x^2 \) and the x-axis from \( x = 0 \) to \( x = 6 \), we need to set up and evaluate the definite integral of the function \( 25 - x^2 \) with respect to \( x \) over the given interval.
The definite integral for this problem is:
\[
\int_{0}^{6} (25 - x^2) \, dx
\]
### Step-by-Step Solution:
1. **Set up the Integral:**
\[
A = \int_{0}^{6} (25 - x^2) \, dx
\]
2. **Integrate the Function:**
\[
\int (25 - x^2) \, dx = 25x - \frac{x^3}{3}
\]
3. **Evaluate the Integral at the Bounds:**
\[
A = \left[ 25x - \frac{x^3}{3} \right]_{0}^{6}
\]
4. **Calculate the Values:**
\[
A = \left(25(6) - \frac{6^3}{3}\right) - \left(25(0) - \frac{0^3}{3}\right)
\]
\[
= \left(150 - 72\right) - (0)
\]
\[
= 78
\]
Thus, the total area is:
\[
\boxed{78}
\]
### Conclusion:
The total area between the curve \( y = 25 - x^2 \) and the x-axis for \( 0 \leq x \leq 6 \) is \( 78 \) square units.
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