11) Consider F(x) = S*sec t tan t dt a) Integrate to find F as a function of x b) Demonstrate the Second Fundamental Theorem of Calculus by differentiating the results in part (a)
11) Consider F(x) = S*sec t tan t dt a) Integrate to find F as a function of x b) Demonstrate the Second Fundamental Theorem of Calculus by differentiating the results in part (a)
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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Use Fundamental Theorem of Calculus part 1 to answer question 12
![### Calculus Exercise on Evaluating Integrals and the Second Fundamental Theorem of Calculus
#### Problem Statement:
11) Consider the function \( F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \)
a) Integrate to find \( F \) as a function of \( x \).
b) Demonstrate the Second Fundamental Theorem of Calculus by differentiating the result found in part (a).
#### Solution:
##### Part (a): Integration to Determine \( F \)
To find \( F(x) \) as a function of \( x \), we need to evaluate the integral \( \int_{a}^{x} \sec{t} \tan{t} \, dt \).
We know that the derivative of \( \sec{t} \) is \( \sec{t} \tan{t} \). Thus, the integral:
\[ \int \sec{t} \tan{t} \, dt = \sec{t} + C \]
Using this, we can determine:
\[ F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \]
Upon evaluating the integral, we have:
\[ F(x) = \sec(x) - \sec(a) \]
Hence, \( F(x) \) is given by:
\[ F(x) = \sec{x} - \sec{a} \]
where \( a \) is a constant.
##### Part (b): Applying the Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus states that if \( F(x) \) is defined as an integral with a variable upper limit, then the derivative of \( F(x) \) with respect to \( x \) is given by the integrand evaluated at \( x \):
\[ \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \]
From part (a), we have:
\[ F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \]
Taking the derivative of \( F(x) \) with respect to \( x \) gives:
\[ \frac{d}{dx} F(x) = \frac{d}{dx} \left( \int_{a}^{x} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2F1dddbf9e-d0b7-4f05-930c-b55252115549%2Fvprstg3.png&w=3840&q=75)
Transcribed Image Text:### Calculus Exercise on Evaluating Integrals and the Second Fundamental Theorem of Calculus
#### Problem Statement:
11) Consider the function \( F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \)
a) Integrate to find \( F \) as a function of \( x \).
b) Demonstrate the Second Fundamental Theorem of Calculus by differentiating the result found in part (a).
#### Solution:
##### Part (a): Integration to Determine \( F \)
To find \( F(x) \) as a function of \( x \), we need to evaluate the integral \( \int_{a}^{x} \sec{t} \tan{t} \, dt \).
We know that the derivative of \( \sec{t} \) is \( \sec{t} \tan{t} \). Thus, the integral:
\[ \int \sec{t} \tan{t} \, dt = \sec{t} + C \]
Using this, we can determine:
\[ F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \]
Upon evaluating the integral, we have:
\[ F(x) = \sec(x) - \sec(a) \]
Hence, \( F(x) \) is given by:
\[ F(x) = \sec{x} - \sec{a} \]
where \( a \) is a constant.
##### Part (b): Applying the Second Fundamental Theorem of Calculus
The Second Fundamental Theorem of Calculus states that if \( F(x) \) is defined as an integral with a variable upper limit, then the derivative of \( F(x) \) with respect to \( x \) is given by the integrand evaluated at \( x \):
\[ \frac{d}{dx} \left( \int_{a}^{x} f(t) \, dt \right) = f(x) \]
From part (a), we have:
\[ F(x) = \int_{a}^{x} \sec{t} \tan{t} \, dt \]
Taking the derivative of \( F(x) \) with respect to \( x \) gives:
\[ \frac{d}{dx} F(x) = \frac{d}{dx} \left( \int_{a}^{x} \
![**Problem 12:** Evaluate the integral
\[ \int \left(2 - \frac{1}{t^6}\right) \left(\frac{1}{t}\right) \, dt \]
This integral involves multiplying two functions and finding their integral with respect to \( t \). The integral combines polynomial and rational functions, which may simplify upon expansion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2F1dddbf9e-d0b7-4f05-930c-b55252115549%2Fb1gzs18.png&w=3840&q=75)
Transcribed Image Text:**Problem 12:** Evaluate the integral
\[ \int \left(2 - \frac{1}{t^6}\right) \left(\frac{1}{t}\right) \, dt \]
This integral involves multiplying two functions and finding their integral with respect to \( t \). The integral combines polynomial and rational functions, which may simplify upon expansion.
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