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 area under the graph of \( f(x) = x^2 + 5 \) between \( x = 0 \) and \( x = 6 \).
**Solution Explanation:**
To find the area under this curve, we need to calculate the definite integral of the function \( f(x) = x^2 + 5 \) from 0 to 6.
**Steps:**
1. **Set Up the Integral:**
The definite integral from \( x = 0 \) to \( x = 6 \) is given by:
\[
\int_{0}^{6} (x^2 + 5) \, dx
\]
2. **Integrate the Function:**
Evaluate the indefinite integral:
\[
\int (x^2 + 5) \, dx = \frac{x^3}{3} + 5x + C
\]
3. **Evaluate the Definite Integral:**
Substitute the upper and lower limits into the antiderivative:
\[
\left[ \frac{x^3}{3} + 5x \right]_{0}^{6} = \left( \frac{6^3}{3} + 5 \times 6 \right) - \left( \frac{0^3}{3} + 5 \times 0 \right)
\]
4. **Calculate:**
\[
\left( \frac{216}{3} + 30 \right) - (0 + 0) = 72 + 30 = 102
\]
**Conclusion:**
The area under the graph of \( f(x) = x^2 + 5 \) from \( x = 0 \) to \( x = 6 \) is 102 square units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc076a9b0-0448-4725-9806-e2628b347801%2Fd75eae66-ea47-494a-b78b-c5d730f8c4b4%2Fro058c5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area under the graph of \( f(x) = x^2 + 5 \) between \( x = 0 \) and \( x = 6 \).
**Solution Explanation:**
To find the area under this curve, we need to calculate the definite integral of the function \( f(x) = x^2 + 5 \) from 0 to 6.
**Steps:**
1. **Set Up the Integral:**
The definite integral from \( x = 0 \) to \( x = 6 \) is given by:
\[
\int_{0}^{6} (x^2 + 5) \, dx
\]
2. **Integrate the Function:**
Evaluate the indefinite integral:
\[
\int (x^2 + 5) \, dx = \frac{x^3}{3} + 5x + C
\]
3. **Evaluate the Definite Integral:**
Substitute the upper and lower limits into the antiderivative:
\[
\left[ \frac{x^3}{3} + 5x \right]_{0}^{6} = \left( \frac{6^3}{3} + 5 \times 6 \right) - \left( \frac{0^3}{3} + 5 \times 0 \right)
\]
4. **Calculate:**
\[
\left( \frac{216}{3} + 30 \right) - (0 + 0) = 72 + 30 = 102
\]
**Conclusion:**
The area under the graph of \( f(x) = x^2 + 5 \) from \( x = 0 \) to \( x = 6 \) is 102 square units.
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