1. ## Improper Integral (have answer, need clarification)

This is from a an AP Calculus BC 2006 exam.

My question is, why start at t = 2? Why add the initial value?
I wrote my improper integral from 0 to infinity.

2. Originally Posted by DiggOlive
This is from a an AP Calculus BC 2006 exam.

My question is, why start at t = 2? Why add the initial value?
I wrote my improper integral from 0 to infinity.
$y(t)-y(2)=\int_2^t \frac{dy}{dt}dt$ and $y(2)=-3$

3. Well, you are only given the derivatives of $x(t)$ and $y(t)$, so there are an infinite number of possible functions with these derivatives. So you need an initial value condition to determine which function you are dealing with. Usually a problem gives you the initial function of $y(0)=k$, but this case you are given the initial condition is $y(2)=-3$. So to find the horizontal asymptote, you have to find the limit of the function (which will be the integral of the derivative) from $t=2$, because we know at $y(2)=-3$. If you weren't given an initial condition, there'd be an infinite number of functions with these derivatives, and therefore there'd be an infinite amount of respective horizontal asymptotes.

### diggolive

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