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Trying to find proper variables for a drinking game wherein you drink 100 times over the course of 100 minutes, wherein the first drink is taken after 0.25 minutes (15 seconds) and the last one is taken after 2 minutes.

I've approximated an exponential function that allows one to do this, $$f(x) = 1.0212266063154^x*0.24480364931149,$$ $ f(1) \sim 0.2499... $ and $ f(100) \sim 2.00... $

Which I'd translate to, the first drink is taken after $f(1)$, and the $100$th after $f(100)$ minutes. However, when I look at the integral of the function $$ \int_1^{100} \!1.0212266063154^x 0.24480364931149 \, \mathrm{d}x = 83.315638611526 $$

Which means, you'd go through the $100$ drinks in only $83.5$ minutes? am I reading this wrong and/or should I go about this in a different way?

I'm not sure how to include the constraint with the integral.

Edit: Realizing how specific my question is worded, I'm of course happy with a solution to the general question of how to progress with finding an exponential function that goes through points $(x_1, y_1)$ and $(x_2, y_2)$ with a definite integral $$ \int_{x_1}^{x_2} \! ab^x \, \mathrm{d}x = z, $$ for a specific $z$.

Thank you.

Edit: I have changed the range of the integral from $y_1, \ldots, y_2$ to $x_1, \ldots, x_2$. As this was an unfortunate typo. I've also introduced $z$ because it might as well be an independent variable.

DrLime2k10
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  • Don't you want the integral to sum up to 100? So set your function with a variable. Find the specific solution to the integral and solve for the variable. – Moti Jan 06 '16 at 06:17
  • I'm not entirely sure how to proceed with your help. Manipulating or generating functions and finding variables with these kinds of constraints is not something I'm familiar with. – DrLime2k10 Jan 06 '16 at 08:58
  • $y_1=0, y_2=100, b=e, and x_2=100$. Let "a" be a variable and calculate it based on your calculation of the integral ("a" can be taken out of the integral) – Moti Jan 06 '16 at 16:45
  • Drinking 100 times in 100 minutes doesn't sound wise... – Théophile Jan 09 '16 at 14:49

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For your second question, $$\int_{y_1}^{y_2} \! ab^x \, \mathrm{d}x = a\int_{y_1}^{y_2} \! e^{x\log b}\, \mathrm{d}x=\left.a\frac{e^{x\log b}}{\log b}\right|_{y_1}^{y_2}=\frac a{\log b}(b^{y_2}-b^{y_1})$$ You have two parameters to fit two points, so should be able to do it. Finding $b$ will take numeric search. Integrating from $y_1$ to $y_2$ is strange-usually you would integrate over the range of $x$

I don't understand what you are asking for the first question. If you sum the geometric series you get about $84.44$, very close to your integral. Why do you expect the sum or integral to be $100$?

Added in response to comment: Your request to find an increasing function with $f(1)=0.25, f(100)=100, \sum_{i=2}^{99}f(i)=97.75$ is a fine one, but you won't get there with a simple exponential, as you have shown. You can't do an arithmetic progression because the average term would be $\frac 12(2+0.25)=1.125$ and the sum would be too large, but you can do a quadratic. If $f(i)=ai^2+bi+c$ we want $$a+b+c=0.25\\10000a+100b+c=2 \\338350a+5050b+100c=100$$ which has a solution $$a = 1/12936 \\ b = 383/38808 \\ c = 2329/9702$$

Ross Millikan
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  • You are correct. Of course I meant to integrate over the range of x. Apologies for having this typo. I will immediately update my question. – DrLime2k10 Jan 07 '16 at 19:48
  • In regards to my first question. In order to spend 100 minutes in total, wouldn't that mean that the area under the curve should be 100? As y = minutes distributed to the x'th 1/100 ? – DrLime2k10 Jan 07 '16 at 19:53
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    No. You have made your exponential go through the points $(1,0.25)$ and $(100,2)$. There is no reason for the integral or sum (which are very close to each other) to be $100$ If your first point were $(1,.01)$ the area would be smaller, but $f(100)$ would still be $100$. If you want the total duration to be $100$, you need the sum to be $100$. The area would be close to $100$ if you used a straight line, but the exponential sags below the straight line, so the area is less. – Ross Millikan Jan 08 '16 at 04:52
  • The idea is, that I want the total duration to be $100$ minutes, and the first, $f(1) = 0.25$ and the last $f(100) = 2$. Such that $f(2) + \ldots + f(99) = 97.75$. Does this make sense? – DrLime2k10 Jan 09 '16 at 12:08