# Indefinite Integration: Sub

• Feb 8th 2009, 04:36 AM
Macleef
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 $\displaystyle -880e^{- \frac{t}{5}}$ dollars per year. If the machine was originally worth $5,200, how much will it be worth when it is 10 years old? Here's what I did: Let$\displaystyle u = - \frac{t}{5}
$Let$\displaystyle dx = 5 du
\displaystyle = - \int 880e^u \times 5 du\displaystyle = - 4400 e^{- \frac{t}{5}} + c\displaystyle 5200 = -4400e^{0} + c\displaystyle 5200 + 4400 = c\displaystyle c = 9600\displaystyle f(10) = -4400e^{-2} + 9600\displaystyle f(10) = 9004.52$...and I'm sure this is wrong since the value has to be less than the initial value and my answer is greater than that value... Please show and explain to me how to get the right answer! • Feb 8th 2009, 05:22 AM skeeter value of the machine in 10 years ...$\displaystyle 5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$• Feb 8th 2009, 02:53 PM Macleef ... the answer is$1,395.48 and I don't know how to get this
• Feb 8th 2009, 03:01 PM
skeeter
Quote:

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

$\displaystyle 5200 - \int_0^{10} 800e^{-\frac{t}{5}} \, dt = 1741.34$

my mistake ... rate coefficient should be 880 instead of 800.

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