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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![**Calculus Problem: Evaluating a Definite Integral**
**Problem Statement:**
Calculate the integral of the function \( \frac{1}{x^2} \) with respect to \( x \) from 1 to 10. Mathematically, this can be expressed as:
\[ \int_{1}^{10} \frac{1}{x^2} \, dx \]
**Explanation:**
The given integral represents the area under the curve \( \frac{1}{x^2} \) from \( x = 1 \) to \( x = 10 \).
To solve this, we first find the antiderivative (indefinite integral) of \( \frac{1}{x^2} \):
\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C \]
Now, we evaluate the definite integral from 1 to 10 using the Fundamental Theorem of Calculus. This involves substituting the upper and lower limits into the antiderivative and finding the difference:
\[ \int_{1}^{10} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^{10} = -\frac{1}{10} - \left( -\frac{1}{1} \right) \]
\[ = -\frac{1}{10} + 1 \]
\[ = 1 - \frac{1}{10} \]
\[ = \frac{10}{10} - \frac{1}{10} \]
\[ = \frac{9}{10} \]
Thus, the value of the integral is \( \frac{9}{10} \).
**Conclusion:**
By evaluating this definite integral, we have determined that the area under the curve \( \frac{1}{x^2} \) from \( x = 1 \) to \( x = 10 \) is \( \frac{9}{10} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdaf4ab92-a308-4681-9263-937919bd13a5%2Fc1b06868-bf72-473b-b84e-3774eabc103a%2Fj0m669f_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculus Problem: Evaluating a Definite Integral**
**Problem Statement:**
Calculate the integral of the function \( \frac{1}{x^2} \) with respect to \( x \) from 1 to 10. Mathematically, this can be expressed as:
\[ \int_{1}^{10} \frac{1}{x^2} \, dx \]
**Explanation:**
The given integral represents the area under the curve \( \frac{1}{x^2} \) from \( x = 1 \) to \( x = 10 \).
To solve this, we first find the antiderivative (indefinite integral) of \( \frac{1}{x^2} \):
\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C \]
Now, we evaluate the definite integral from 1 to 10 using the Fundamental Theorem of Calculus. This involves substituting the upper and lower limits into the antiderivative and finding the difference:
\[ \int_{1}^{10} \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^{10} = -\frac{1}{10} - \left( -\frac{1}{1} \right) \]
\[ = -\frac{1}{10} + 1 \]
\[ = 1 - \frac{1}{10} \]
\[ = \frac{10}{10} - \frac{1}{10} \]
\[ = \frac{9}{10} \]
Thus, the value of the integral is \( \frac{9}{10} \).
**Conclusion:**
By evaluating this definite integral, we have determined that the area under the curve \( \frac{1}{x^2} \) from \( x = 1 \) to \( x = 10 \) is \( \frac{9}{10} \).
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