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 Statement:**
Find the area under the graph of \( f(x) = x^2 + 10 \) between \( x = 0 \) and \( x = 6 \).
**Solution:**
To find the area under the curve of the function \( f(x) = x^2 + 10 \) from \( x = 0 \) to \( x = 6 \), we need to calculate the definite integral of the function with respect to \( x \) over the interval [0, 6].
The definite integral is given by:
\[
\int_{0}^{6} (x^2 + 10) \, dx
\]
Calculate the integral:
1. Integrate \( x^2 \):
\[
\int x^2 \, dx = \frac{x^3}{3}
\]
2. Integrate the constant 10:
\[
\int 10 \, dx = 10x
\]
3. Combine the integrals:
\[
\int (x^2 + 10) \, dx = \frac{x^3}{3} + 10x
\]
4. Evaluate the definite integral from 0 to 6:
\[
\left[ \frac{x^3}{3} + 10x \right]_{0}^{6} = \left( \frac{6^3}{3} + 10 \times 6 \right) - \left( \frac{0^3}{3} + 10 \times 0 \right)
\]
5. Simplify the expression:
\[
= \left( \frac{216}{3} + 60 \right) - (0 + 0)
\]
\[
= (72 + 60)
\]
\[
= 132
\]
**Conclusion:**
The area under the curve of \( f(x) = x^2 + 10 \) from \( x = 0 \) to \( x = 6 \) is 132 square units.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe59facd3-ea8e-4c1d-9c3a-436e1d3f8729%2Fc4793881-23a7-4d30-a6ab-b6d8394b97c6%2F8e0ndzu_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the area under the graph of \( f(x) = x^2 + 10 \) between \( x = 0 \) and \( x = 6 \).
**Solution:**
To find the area under the curve of the function \( f(x) = x^2 + 10 \) from \( x = 0 \) to \( x = 6 \), we need to calculate the definite integral of the function with respect to \( x \) over the interval [0, 6].
The definite integral is given by:
\[
\int_{0}^{6} (x^2 + 10) \, dx
\]
Calculate the integral:
1. Integrate \( x^2 \):
\[
\int x^2 \, dx = \frac{x^3}{3}
\]
2. Integrate the constant 10:
\[
\int 10 \, dx = 10x
\]
3. Combine the integrals:
\[
\int (x^2 + 10) \, dx = \frac{x^3}{3} + 10x
\]
4. Evaluate the definite integral from 0 to 6:
\[
\left[ \frac{x^3}{3} + 10x \right]_{0}^{6} = \left( \frac{6^3}{3} + 10 \times 6 \right) - \left( \frac{0^3}{3} + 10 \times 0 \right)
\]
5. Simplify the expression:
\[
= \left( \frac{216}{3} + 60 \right) - (0 + 0)
\]
\[
= (72 + 60)
\]
\[
= 132
\]
**Conclusion:**
The area under the curve of \( f(x) = x^2 + 10 \) from \( x = 0 \) to \( x = 6 \) is 132 square units.
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