How do you find the integral of acceleration I'm doing online physics homework. Velocity is the integral of acceleration. I understand that, but for whatever reason my calculations aren't working. $$ax(t) = 4t - 2t^2 + 45e^{-t/F}$$ where $F$ is a constant $1$ second and the initial velocity is $15$. When I try to find the integral, I get $$vx(t)= -2t^3/3 + 2t^2 -45e^{-1} + 15$$ I plug in the seconds, and it doesn't work. $1$ second somehow equals $44.78$ and $2$ seconds equal $56.58$, but those aren't the numbers i'm getting. Now i'm trying to find velocity at $3$ second. I think $45e^{-t}$ may be the root of the problem, but I have no clue what to do about it. 
 A: 
When I try to find the integral, I get $$vx(t)= -2t^3/3 + 2t^2 -45e^{-1} + 15$$

The main problem is that you made a mistake in integrating
$$\int 45e^{-t/F}dt=-45Fe^{-t/F}+C$$
by which you should redo the integration to arrive at the correct answer. Given that the acceleration is
$$a(t) = 4t - 2t^2 + 45e^{-t/F}$$
we know that 
$$v(t)=\int a(t)\,dt$$
therefore
\begin{align}v(t)&=\int\left(4t - 2t^2 + 45e^{-t/F}\right)dt&\\&=\frac{4t^2}{2}-\frac{2t^3}{3}-45Fe^{-t/F}+C\\&=
2t^2-\frac{2t^3}{3}-45Fe^{-t/F}+C
\end{align}
from which we can evaluate the initial velocity by setting $t=0$
$$v(0)=0-0-45F+C=15$$
and since $F=1$ this means that
$$C=60$$
so that
$$v(t)=2t^2-\frac{2t^3}{3}-45Fe^{-t/F}+60$$
from which
$$v(1)\approx 44.78$$
$$v(2)\approx 56.58$$
Remark 1: The units of acceleration are $m/s^2$ (meters per seconds squared) and the units for velocity are $m/s$ (meters per second). Since this is a physics question, I would include the units in your answer.
Remark 2: I don't know why you wrote
$$ax(t)$$
as the acceleration function and 
$$vx(t)$$
as the velocity function. This is nonstandard and I would follow the notation used in your course notes (or textbook). The standard that I know of is $a(t)$ for the acceleration function, $v(t)$ for the velocity function, and $s(t)$ for the position function.
A: We have
$$a_x(t) = 4t - 2t^2 + 45e^{-t/F}\implies v_x(t)=2t^2-\frac23t^3-45Fe^{-t/F}+c$$
with $$v(0)=-45+c=15\implies c=60$$
and therefore 
$$v_x(1)=2-\frac23-45/e+60\approx 44.78$$
