Solving Reverse Integrals: Find f(x) to Solve 1-0.1^n

[Big-Daddy]

I need to find a function f(x) such that

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

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 ...

 

[mathman]

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]

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]

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) = .