Mth 272
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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Mr. Smith,
I am working on assignment 16 trying to understand example 7 on page 425 (eight edition) I am not seeing how v = -20 e^-0.05t.
I would take intergal of e^-0.05t dt let u = -.0.05 I believe you meant -.05 t; probably just a typo that the t wasn't included
and du/dt = -0.05 which would be e^u * du/dx = e^-0.05t(-0.05) = -0.05e^(-0.05t).
Not understanding where or how they got -20.
Tim
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Good question. Hopefully my answer will help clarify some important aspects of integration by substitution.
The derivative of -.05 * e^(-.05 t) is -.05 * ( -.05 e^(-.05 t)) = .0025 e^(-.05 t). This isn't equal to the original integrand, so your integration has an error.
IMPORTANT NOTE: Always take the derivative of your antiderivative to check your integration. It's always worth the time and effort, an greatly accelerates the process of learning integration technique by connecting the process to the result.
TO integrate e^-0.05t dt you are right to let u = -.0.05 t so that du/dt = -0.05.
This gives you du = -.05 dt.
The integral int(e^(-.05 t) dt) is in terms of dt, so you need an expression for dt. Since du = -.05 dt, we have dt = du / (-.05), so that our integral becomes
int(e^u * du / (-.05) ).
Since 1/(-.05) = -20, this can be rewritten (using the constant rule) as
-20 int(e^u du), which is just
-20 * e^u + c. Substituting for u our integral becomes
-20 e^(-.05 t) + c.
You did give me all the information I needed to answer your question, provided I've interpreted your question correctly.
For future reference note that I can't usually respond to a question which refers me to a problem from the text. It's usually important for you to include a brief statement of the problem. I very seldom refer to a textbook, and I have the textbook in only one location.