Integrating 110*1.10^(t) from 0 to 35: Solution

[beanryu]

What is the integral of 110*1.10^(t) from b=35 to a=0 (b being hte upper number to the integral sign and a being the lower number to the integral sign)

From the textbook it says it's the constant c=110 times the sum of the integral of b - the integral of a.

so how do i get integral of a and b?

Thank you for replying!

 

[Benny]

You could try to make use of $$\exp \left( {t\log \left( {1.10} \right)} \right) = \exp \left( {\log \left( {1.10} \right)^t } \right) = 1.10^t$$.

 

[beanryu]

Thanx Benny~! Thank You!

 

[beanryu]

HEY DeAR BENNY
I got another problem if you could help again... I would really really appreciate it

10+990e^(-0.1t)
now to so far i know its antiderivative starts with

10t+ ... something... could you guys give me some hint as to how to continue?

 

[Benny]

If k is a constant then $$\int {e^{kt} } dt = \frac{1}{k}e^{kt} + c$$ where c is an arbitrary constant. Just use that to integrate the exponential. By the way, I would have thought that this one would have been very simple for you if you were able to do the first problem you asked about.