Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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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.
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**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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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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