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Magazine Chapter 17: Problem 6
Find \(\int t \mathrm{e}^{-s t} \mathrm{~d} t\) where \(s\) is a constant.
Short Answer
Expert verified
The integral is \( \int t \mathrm{e}^{-st} \, \mathrm{d}t = -\frac{t}{s} \mathrm{e}^{-st} - \frac{1}{s^2} \mathrm{e}^{-st} + C \).
Step by step solution
01
Identify Integration Method
The integrand is the product of a polynomial function \(t\) and an exponential function \(\mathrm{e}^{-st}\). The integration technique called "Integration by Parts" is suitable here.
02
Set Up Integration by Parts
Integration by parts is based on the formula: \[ \int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u \]For our integral, let \( u = t \) and \( \mathrm{d}v = \mathrm{e}^{-st} \, \mathrm{d}t \).
03
Differentiate and Integrate Parts
Calculate \( \mathrm{d}u \) and \( v \): - \( \mathrm{d}u = \mathrm{d}t \)- To find \( v \), integrate \( \mathrm{d}v \): \[ v = \int \mathrm{e}^{-st} \, \mathrm{d}t = -\frac{1}{s} \mathrm{e}^{-st} \]
04
Apply Integration by Parts Formula
Substitute \( u, \mathrm{d}u, v, \) and \( \mathrm{d}v \) into the integration by parts formula: \[ \int t \mathrm{e}^{-st} \, \mathrm{d}t = t \left(-\frac{1}{s} \mathrm{e}^{-st}\right) - \int \left(-\frac{1}{s} \mathrm{e}^{-st}\right) \, \mathrm{d}t \]This simplifies to: \[ -\frac{t}{s} \mathrm{e}^{-st} + \frac{1}{s} \int \mathrm{e}^{-st} \, \mathrm{d}t \]
05
Evaluate Remaining Integral
Compute the remaining integral: \[ \int \mathrm{e}^{-st} \, \mathrm{d}t = -\frac{1}{s} \mathrm{e}^{-st} \]Thus, substitute back: \[ \int t \mathrm{e}^{-st} \, \mathrm{d}t = -\frac{t}{s} \mathrm{e}^{-st} - \frac{1}{s^2} \mathrm{e}^{-st} + C \]where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Integration
In calculus, exponential integration is a crucial technique for solving integrals involving exponential functions such as \( \mathrm{e}^{-st} \), where \( s \) is a constant. Understanding how to handle these functions is key.
- **Exponential Function Basics:** An exponential function is of the form \( f(t) = a\mathrm{e}^{bt} \), where \( a \) and \( b \) are constants, and \( \mathrm{e} \) is Euler鈥檚 number. This function is unique because its derivative remains proportional to the function itself.
- **Integration Strategy:** Integrating exponential functions often involves recognizing the function type and using basic integration rules. For instance, \( \int \mathrm{e}^{bt} \mathrm{d}t = \frac{1}{b}\mathrm{e}^{bt} + C \).
- **Negative Signs in Exponents:** Special attention should be given to the sign in the exponent. If the exponent is negative, as in \( \mathrm{e}^{-st} \), the integral becomes \( -\frac{1}{s}\mathrm{e}^{-st} + C \).
Polynomial Function Integration
Polynomial functions in calculus are among the simplest to integrate, yet they are ubiquitous in both practice and theory. Recognizing when you're dealing with a polynomial is straightforward.
- **Understanding Polynomials:** A polynomial function can be described as \( f(t) = a_n t^n + a_{n-1} t^{n-1} + ... + a_1 t + a_0 \). The coefficients \( a \) and the powers of \( t \) determine the distinct characteristics of polynomials.
- **Integration Process:** The integral of \( t^n \) is found using the rule \( \int t^n \mathrm{d}t = \frac{t^{n+1}}{n+1} + C \), where \( C \) is the constant of integration. Simply increase the power by one and divide by that new power.
- **Handling Constants:** In integrals featuring constants like \( \int t \cdot \mathrm{e}^{-st} \), you treat \( t \) as the variable and apply standard polynomial integration rules, even when combined with other functions.
Calculus Techniques
Calculus offers a variety of techniques for solving integrals by simplifying the expressions involved. One of the powerful methods is **Integration by Parts**.
- **Integration by Parts Formula:** This technique is derived from the product rule for differentiation and is formulated as \( \int u \, \mathrm{d}v = uv - \int v \, \mathrm{d}u \). It's particularly useful when an integrand is a product of a function easy to differentiate (2u2) and a function easy to integrate (2d2v).
- **Choosing \( u \) and \( \mathrm{d}v \):** The approach typically involves choosing \( u \) to be the polynomial part, such as \( t \), because differentiating it simplifies the expression. For \( \mathrm{d}v \), you select the exponential portion, like \( \mathrm{e}^{-st} \, \mathrm{d}t \).
- **Solving Integrals with Parts:** Once you've decided on \( u \) and \( \mathrm{d}v \), differentiate \( u \) and integrate \( \mathrm{d}v \), then substitute into the integration by parts formula. This will transform the integral into a simpler form to evaluate.
