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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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
Linear Functions
A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
Question
![**Problem: Compute the exact value**
\[ \int_{2}^{\infty} 2^{-x} \, dx \]
There are no graphs or diagrams associated with this problem. The goal is to compute the exact value of the improper integral from 2 to infinity of the function \(2^{-x}\).
### Solution Outline:
1. **Identify the Integrand**: The integrand is \(2^{-x}\), which can be written as \(e^{-x \ln(2)}\) using the change of base formula.
2. **Perform the Integration**:
\[
\int 2^{-x} \, dx = \int e^{-x \ln(2)} \, dx
\]
Let \(k = \ln(2)\), then the integral becomes:
\[
\int e^{-kx} \, dx
\]
3. **Integrate the Exponential Function**:
The integral of \(e^{-kx}\) with respect to \(x\) is:
\[
\int e^{-kx} \, dx = -\frac{1}{k} e^{-kx} + C
\]
Substituting \(k = \ln(2)\):
\[
\int 2^{-x} \, dx = -\frac{1}{\ln(2)} 2^{-x} + C
\]
4. **Evaluate the Definite Integral**:
Evaluate the definite integral from 2 to infinity:
\[
\left[ -\frac{1}{\ln(2)} 2^{-x} \right]_{2}^{\infty}
\]
5. **Evaluate the Limits**:
\[
= \left( \lim_{{t \to \infty}} \left( -\frac{1}{\ln(2)} 2^{-t} \right) \right) - \left( -\frac{1}{\ln(2)} 2^{-2} \right)
\]
As \(t \to \infty\), \(2^{-t} \to 0\):
\[
= 0 - \left( -\frac{1}{\ln(2)} 2^{-2} \right)
\]
Simplify the expression:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c8732dd-627e-495c-a245-47d9f524edaa%2F27afcf13-df07-4143-a23a-f149bba20805%2Fni15c4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem: Compute the exact value**
\[ \int_{2}^{\infty} 2^{-x} \, dx \]
There are no graphs or diagrams associated with this problem. The goal is to compute the exact value of the improper integral from 2 to infinity of the function \(2^{-x}\).
### Solution Outline:
1. **Identify the Integrand**: The integrand is \(2^{-x}\), which can be written as \(e^{-x \ln(2)}\) using the change of base formula.
2. **Perform the Integration**:
\[
\int 2^{-x} \, dx = \int e^{-x \ln(2)} \, dx
\]
Let \(k = \ln(2)\), then the integral becomes:
\[
\int e^{-kx} \, dx
\]
3. **Integrate the Exponential Function**:
The integral of \(e^{-kx}\) with respect to \(x\) is:
\[
\int e^{-kx} \, dx = -\frac{1}{k} e^{-kx} + C
\]
Substituting \(k = \ln(2)\):
\[
\int 2^{-x} \, dx = -\frac{1}{\ln(2)} 2^{-x} + C
\]
4. **Evaluate the Definite Integral**:
Evaluate the definite integral from 2 to infinity:
\[
\left[ -\frac{1}{\ln(2)} 2^{-x} \right]_{2}^{\infty}
\]
5. **Evaluate the Limits**:
\[
= \left( \lim_{{t \to \infty}} \left( -\frac{1}{\ln(2)} 2^{-t} \right) \right) - \left( -\frac{1}{\ln(2)} 2^{-2} \right)
\]
As \(t \to \infty\), \(2^{-t} \to 0\):
\[
= 0 - \left( -\frac{1}{\ln(2)} 2^{-2} \right)
\]
Simplify the expression:
\[
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