Business: Total Savings. A company installs a new computer that is expected to generate savings at the rate 20,000e^-0.02t dollars per year, where t is the number of years that the computer has been in operation. If the computer originally cost $230,000, when will it "pay for itself"? Round answers to the nearest tenth. [Hint: Find the indefinite integral, solve for the constant C.] I keep coming down to the end of it (or so it seems) and getting e^-.02t+C=-.23 I know I'm doing something wrong, I just can't figure out where. Thanks in advance. 2. Originally Posted by FGCUguy Business: Total Savings. A company installs a new computer that is expected to generate savings at the rate 20,000e^-0.02t dollars per year, where t is the number of years that the computer has been in operation. If the computer originally cost$230,000, when will it "pay for itself"? Round answers to the nearest tenth. [Hint: Find the indefinite integral, solve for the constant C.]
So you need to find when the sum of all savings, given by $\displaystyle s(t)=20,000e^{-.02t}$, is equal to $230k. You are right that integration is the way to do this, since you need to account for the total savings, not just the savings at a particular time. Your problem is this:$\displaystyle 230,000 = \int_{0}^{x}20,000e^{-.02t}dt$You want to find x, which is the time in years it took to reach the initial investment. That is how I would go about it, but that uses definite integration instead of indefinite. Does this make sense to you? edit: The equation you gave for the savings indicates that at t=0, the savings is already$20k because e^0 is 1. If that is true, then I suppose you need to find when the integral equals $210,000. I'll think about this some more. Didn't want to give you an incorrect answer though. 3. Ok I think I got it now doing it your way. If you integrate the savings equation with respect to time you get$\displaystyle \int s(t)dt +C$. Firstly, show us all your work because I don't know how you got to your final answer that you gave. Anyway, the integral of s(t) will input t and output the total amount saved. This is good and what we want. However we need to find C. How to do this? Well use the fact that at t=0, when the new computer was installed, that the savings was 0. This is why I was confused earlier, because at t=0, the rate of savings isn't 0. So you need to solve$\displaystyle \int s(t)dt + C = 0$, integrating properly and plugging in t=0, to indicate that the computer was just bought. Then you can solve for C, then you can find when$\displaystyle \int s(t) dt + C = 230,000$4. Sorry in advance, I don't know the code for all the symbols... But here's my work... Well assume int. is the same thing as the integral sign. int.20,000e^-.02t dt 1/-.02(20000)int.e^-.02t(-.02)dt -1000000int.e^-.02t(-.02)dt -1000000int.e^tdt -1000000[e^-.02t+C]/-1000000 = 230000/-1000000 e^-.02t+C=-.23 hope thats not too much of a mess to understand. 5. Originally Posted by FGCUguy Sorry in advance, I don't know the code for all the symbols... But here's my work... Well assume int. is the same thing as the integral sign. int.20,000e^-.02t dt 1/-.02(20000)int.e^-.02t(-.02)dt -1000000int.e^-.02t(-.02)dt -1000000int.e^tdt -1000000[e^-.02t+C]/-1000000 = 230000/-1000000 e^-.02t+C=-.23 hope thats not too much of a mess to understand. You're making this integral harder than it is. The coefficient is right. You have to divide by the coefficient of e, which is 1,000,000 like you wrote. However the$\displaystyle \int e^{-x}dx = \frac{1}{-1}e^{-x}+C$. I wrote a simplified version of your problem. The "e" term stays the same. I don't know what you're doing with the parts in red. Also, all of a sudden you start with an integral and move to an equation. What happened there? Where are these numbers coming from?$\displaystyle \int 20,000 e^{-.02t}dt = -1,000,000e^{-.02t}+C\$