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Reverse integrals

by Big-Daddy
Tags: integrals, reverse
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Big-Daddy
#1
Jan14-14, 02:54 PM
P: 343
I need to find a function f(x) such that

[tex]\int_{-\infty}^{100+10n} (f(x)) dx = 1-0.1^n[/tex]

for n=1,2,3,4,5,6...∞. How would I go about this? It must be exponential in some way I'm guessing?

This is not a homework problem. I don't just want the answer. I want guidance on this type of problem and function, but please from someone with an idea of how to answer this particular case too ...
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mathman
#2
Jan14-14, 03:12 PM
Sci Advisor
P: 6,059
Replace (100+10n) by a real variable (y). Take the derivative of both sides, this will give you an expression for f(y) if it exists. It will be an exponential.
Big-Daddy
#3
Jan14-14, 05:18 PM
P: 343
Quote Quote by mathman View Post
Replace (100+10n) by a real variable (y). Take the derivative of both sides, this will give you an expression for f(y) if it exists. It will be an exponential.
So y=100+10n, n = 1/10(y-100) = (1/10)y-10.

Now what? I need to bear the integral limits (-∞ to y) in mind...

mathman
#4
Jan15-14, 03:20 PM
Sci Advisor
P: 6,059
Reverse integrals

0.1n = 0.1(y-100)/10 = exp((y-100)ln(0.1)/10). Now take the derivatives of both sides to get f(y) = .


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