# Thread: Indefinite Integration: Sub

1. ## Indefinite Integration: Sub

The resale value of a certain industrial machine decreases at a rate that depends on its age. When the machine is t years old, the rate at which its value is changing is $-880e^{- \frac{t}{5}}$ dollars per year. If the machine was originally worth $5,200, how much will it be worth when it is 10 years old? Here's what I did: Let $u = - \frac{t}{5} $ Let $dx = 5 du $ $= - \int 880e^u \times 5 du$ $= - 4400 e^{- \frac{t}{5}} + c$ $5200 = -4400e^{0} + c$ $5200 + 4400 = c$ $c = 9600$ $f(10) = -4400e^{-2} + 9600$ $f(10) = 9004.52$ ...and I'm sure this is wrong since the value has to be less than the initial value and my answer is greater than that value... Please show and explain to me how to get the right answer! 2. value of the machine in 10 years ... $5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$ 3. ... the answer is$1,395.48 and I don't know how to get this

4. Originally Posted by skeeter
value of the machine in 10 years ...

$5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$
my mistake ... rate coefficient should be 880 instead of 800.

$5200 - \int_0^{10} 880e^{-\frac{t}{5}} \, dt = 1395.48$