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
Please use Fundamental theorem of calculus for #8 and answer both questions thank you so much!
![**Problem 8:** Find the area of the region bounded by the graphs of the equation \( f(x) = \frac{2}{x} \), \( y = 0 \), \( x = 1 \), and \( x = 6 \).
This problem requires determining the area under the curve defined by the function \( f(x) = \frac{2}{x} \) from \( x = 1 \) to \( x = 6 \), and above the x-axis (\( y = 0 \)). This problem involves integrating the function \( f(x) \) over the interval \([1, 6]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2F22b6aef5-cd78-4a0f-bf4f-fc8132a82aef%2F3f39w3c.png&w=3840&q=75)
Transcribed Image Text:**Problem 8:** Find the area of the region bounded by the graphs of the equation \( f(x) = \frac{2}{x} \), \( y = 0 \), \( x = 1 \), and \( x = 6 \).
This problem requires determining the area under the curve defined by the function \( f(x) = \frac{2}{x} \) from \( x = 1 \) to \( x = 6 \), and above the x-axis (\( y = 0 \)). This problem involves integrating the function \( f(x) \) over the interval \([1, 6]\).
![## Mean Value Theorem for Integrals: Example Problem
### Problem Statement:
9) Find the value(s) of \( c \) guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.
\[ f(x) = 2 \sin x \]
\[ \left[ 0, \frac{\pi}{3} \right] \]
### Solution Process:
To apply the Mean Value Theorem for Integrals, which states that if \( f \) is continuous on \([a, b]\), then there exists a number \( c \) in \((a, b)\) such that:
\[ \int_a^b f(x) \, dx = f(c) (b - a) \]
First, find the definite integral of \( f(x) \) over the interval \([0, \frac{\pi}{3}]\).
\[ \int_0^{\frac{\pi}{3}} 2 \sin x \, dx \]
Next, find the antiderivative of \( 2 \sin x \):
\[ 2 \int_0^{\frac{\pi}{3}} \sin x \, dx \]
\[
= 2 \left[ -\cos x \right]_0^{\frac{\pi}{3}}
\]
Plug in the bounds of the interval:
\[
= 2 \left[ -\cos \left( \frac{\pi}{3} \right) + \cos (0) \right]
\]
Evaluate the trigonometric functions:
\[
= 2 \left[ -\frac{1}{2} + 1 \right]
\]
\[
= 2 \left[ \frac{1}{2} \right]
\]
\[
= 1
\]
So, the integral evaluates to 1.
By the Mean Value Theorem for Integrals:
\[
\int_0^{\frac{\pi}{3}} 2 \sin x \, dx = f(c) \left( \frac{\pi}{3} - 0 \right)
\]
We know the integral is 1, so set up the equation:
\[
1 = 2 \sin(c) \left( \frac{\pi}{3} \right)
\]
Solve for \( \sin(c) \):
\[
1 = \frac{2\pi}{3} \sin(c)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a908816-dcaf-4aa4-8dd6-7dd2f0aea84f%2F22b6aef5-cd78-4a0f-bf4f-fc8132a82aef%2Fajdxe8k.png&w=3840&q=75)
Transcribed Image Text:## Mean Value Theorem for Integrals: Example Problem
### Problem Statement:
9) Find the value(s) of \( c \) guaranteed by the Mean Value Theorem for Integrals for the function over the given interval.
\[ f(x) = 2 \sin x \]
\[ \left[ 0, \frac{\pi}{3} \right] \]
### Solution Process:
To apply the Mean Value Theorem for Integrals, which states that if \( f \) is continuous on \([a, b]\), then there exists a number \( c \) in \((a, b)\) such that:
\[ \int_a^b f(x) \, dx = f(c) (b - a) \]
First, find the definite integral of \( f(x) \) over the interval \([0, \frac{\pi}{3}]\).
\[ \int_0^{\frac{\pi}{3}} 2 \sin x \, dx \]
Next, find the antiderivative of \( 2 \sin x \):
\[ 2 \int_0^{\frac{\pi}{3}} \sin x \, dx \]
\[
= 2 \left[ -\cos x \right]_0^{\frac{\pi}{3}}
\]
Plug in the bounds of the interval:
\[
= 2 \left[ -\cos \left( \frac{\pi}{3} \right) + \cos (0) \right]
\]
Evaluate the trigonometric functions:
\[
= 2 \left[ -\frac{1}{2} + 1 \right]
\]
\[
= 2 \left[ \frac{1}{2} \right]
\]
\[
= 1
\]
So, the integral evaluates to 1.
By the Mean Value Theorem for Integrals:
\[
\int_0^{\frac{\pi}{3}} 2 \sin x \, dx = f(c) \left( \frac{\pi}{3} - 0 \right)
\]
We know the integral is 1, so set up the equation:
\[
1 = 2 \sin(c) \left( \frac{\pi}{3} \right)
\]
Solve for \( \sin(c) \):
\[
1 = \frac{2\pi}{3} \sin(c)
Expert Solution
Question - 8
Shaded area is our required area .
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