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Chapter 17: Problem 6

Find \(\int t \mathrm{e}^{-s t} \mathrm{~d} t\) where \(s\) is a constant.

Short Answer

Expert verified
Question: Find the integral of the function \(t \mathrm{e}^{-st}\) with respect to \(t\), where \(s\) is a constant. Answer: The integral of the function \(t \mathrm{e}^{-st}\) with respect to \(t\) is \(-\frac{1}{s}te^{-st} + \frac{1}{s^2} e^{-st} + C\), where \(C\) is the constant of integration.

Step by step solution

01

Choose u and dv

For integration by parts, we have to choose u and dv. In this problem, we will choose: $$ u = t \qquad (\mathrm{d}u = \mathrm{d}t), \quad \mathrm{d}v = e^{-st} \mathrm{d}t \qquad (v = -\frac{1}{s}e^{-st}). $$
02

Apply Integration by Parts Formula

Now, we will apply the integration by parts formula using our chosen u, v, du, and dv: $$ \int t e^{-st} \mathrm{d}t = t(-\frac{1}{s}e^{-st}) - \int (-\frac{1}{s}e^{-st}) \mathrm{d}t. $$
03

Solve the Remaining Integral

Simplify the expression and solve the remaining integral: $$ \int t e^{-st} \mathrm{d}t = -\frac{1}{s}te^{-st} + \frac{1}{s} \int e^{-st} \mathrm{d}t. $$ The remaining integral can be solved using a simple integration rule. The integral of \(e^{at}\) is \(\frac{1}{a}e^{at} + C\), so we have: $$ \int e^{-st} \mathrm{d}t = -\frac{1}{s}e^{-st} + C. $$
04

Substitute and Simplify

Now, substitute the result of Step 3 into the equation from Step 2 and simplify: $$ \int t e^{-st} \mathrm{d}t = -\frac{1}{s}te^{-st} + \frac{1}{s^2} e^{-st} + C. $$ This is the final expression for the integral of the given function with respect to \(t\).

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