Changing variable of integration for int[0,infty] xL e^{xL} dx (using y = xL)
- Thread starter regman900
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Please try to work through step-by-step with writing on paper. If you are still stuck - please come back and show your work.
\(\displaystyle x = \frac{y}{\lambda} \)
\(\displaystyle \int_{0}^{\infty} \lambda \frac{y}{\lambda} e^{-\lambda \frac{y}{\lambda} } d\frac{y}{\lambda} = \int_{0}^{\infty} ye^{-y}d\frac{y}{\lambda}\)
I'm not sure where to go from here, or if even this is correct. I've never seen this technique before and can't find a reference to it anywhere. It looks similar to u substitution, but is obviously different.
\(\displaystyle y = \lambda \frac{y}{\lambda} \)
\(\displaystyle \int_{0}^{\infty} \lambda \frac{y}{\lambda} e^{-\lambda \frac{y}{\lambda}} d\frac{y}{\lambda} \lambda = \int_{0}^{\infty} y e^{-y } dy\)
\(\displaystyle \int_{0}^{\infty} \lambda \frac{y}{\lambda} e^{-\lambda \frac{y}{\lambda} } d\frac{y}{\lambda} = \int_{0}^{\infty} ye^{-y}d\frac{y}{\lambda}\)
I'm not sure where to go from here, or if even this is correct. I've never seen this technique before and can't find a reference to it anywhere. It looks similar to u substitution, but is obviously different.
\(\displaystyle y = \lambda \frac{y}{\lambda} \)
\(\displaystyle \int_{0}^{\infty} \lambda \frac{y}{\lambda} e^{-\lambda \frac{y}{\lambda}} d\frac{y}{\lambda} \lambda = \int_{0}^{\infty} y e^{-y } dy\)
\(\displaystyle \int_{0}^{\infty} x\lambda e^{-\lambda x} dx
= \lambda x(- \frac{1}{\lambda})e^{-\lambda x} - \int_{0}^{\infty} \lambda(-\frac{1}{\lambda})e^{-\lambda x}dx \)
This is integration by parts, but can't find the proper latex syntax for it. I hope you understand.
\(\displaystyle = -xe^{ \lambda x} - \frac{1}{\lambda}e{-\lambda x} = \lambda\)
\(\displaystyle \frac{1}{\lambda} \int_{0}^{\infty} ye^{-y}dy = \frac{1}{\lambda} (y(-e^{-y}) - \int_{0}^{\infty} -e^{-y}dy) = {\lambda}\)
So both are equivalent in this instance. But how do we know before hand? I'm assuming this is a general rule. Does it have a name? Does it have a proof?
= \lambda x(- \frac{1}{\lambda})e^{-\lambda x} - \int_{0}^{\infty} \lambda(-\frac{1}{\lambda})e^{-\lambda x}dx \)
This is integration by parts, but can't find the proper latex syntax for it. I hope you understand.
\(\displaystyle = -xe^{ \lambda x} - \frac{1}{\lambda}e{-\lambda x} = \lambda\)
\(\displaystyle \frac{1}{\lambda} \int_{0}^{\infty} ye^{-y}dy = \frac{1}{\lambda} (y(-e^{-y}) - \int_{0}^{\infty} -e^{-y}dy) = {\lambda}\)
So both are equivalent in this instance. But how do we know before hand? I'm assuming this is a general rule. Does it have a name? Does it have a proof?
Dr.Peterson
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The key idea is that \(\displaystyle dy = \lambda dx \) . Do you see how that comes in?
I do see that. But I don't see how the lambda fraction applies. Isn't the dy purely notional? If so how can the lambda fraction effect it? How can we infer from knowing that \(\displaystyle dy = /lambda dx \) to knowing that dividing the entire integral by \(\displaystyle \frac{1}{\lambda} [\tex] will make it equivalent? Does the technique have a name? Is there a proof anywhere?\)
When you substitute y for x, you must substitute dy for dx and change the limits of integration. But substituting dy for dx requires finding out what the relationship between dy and dx is.
Finally figured it out. It is just a basic U substitution.
\(\displaystyle f(x) = e^{-x} \)
\(\displaystyle g(x) = x\lambda \)
\(\displaystyle g'(x) = \lambda dx \)
\(\displaystyle \int_{0}^{\infty} x\lambda e^{-x\lambda} dx = \frac{1}{\lambda} \int_{0}^{\infty} xe^{-x\lambda}\lambda dx\)
The \(\displaystyle dy = \lambda dx \) really helped. Thanks a lot everyone.
\(\displaystyle f(x) = e^{-x} \)
\(\displaystyle g(x) = x\lambda \)
\(\displaystyle g'(x) = \lambda dx \)
\(\displaystyle \int_{0}^{\infty} x\lambda e^{-x\lambda} dx = \frac{1}{\lambda} \int_{0}^{\infty} xe^{-x\lambda}\lambda dx\)
The \(\displaystyle dy = \lambda dx \) really helped. Thanks a lot everyone.