The function $A$ is given by the formula $f(t)=4-t$ and integral from $A(x) = \int_0^x f(t) dt$ Sketch a graph of $y=A(x)$ for $0≤x≤4$.  
Calculate the values of $A(0)$, $A(1)$, $A(2)$, and $A(3)$.
Determine the values of $A'(0)$, $A'(1)$, $A'(2)$, and $A'(3)$. 
So I know the indefinite integral of $f(t)$ is $$4t - \frac{1}{2}t^{2} + c$$
But I'm not really sure what the next step is.  Any help would be appreciated.
 A: By definition,
$$A(x) = \int_{0}^{x} f(t)~dt$$
where $f(t) = 4 - t$.  Evaluating the integral yields
\begin{align*}
A(x) & = \int_{0}^{x} (4 - t)~dt\\
     & = \left(4t - \frac{1}{2}t^2\right) \bigg|_{0}^{x}\\
     & = 4x - \frac{1}{2}x^2 - (0 - 0)\\
     & = 4x - \frac{1}{2}x^2
\end{align*}
Can you take it from there?
A: Well, then $$A(x) = \left[4t - \frac{1}{2}t^2\right]_0^x$$
$$A(x) = 4x - \frac{1}{2}x^2$$
$$A(x) = \frac{1}{2}\left(8x - x^2\right)$$
$$A(x) = \frac{1}{2}x\left(8 - x\right)$$
The graph of $A(x)$ will be a downward parabola cutting the x-axis at $0$ and $8$.
$$A'(x) = \frac{1}{2}\left(8 - 2x\right) = 4 - x$$  
A: Your indefinite integral is correct but the integral you are using is definite. The constants are not needed as they cancel out.
Note that the definite integral has the formula $\int_{a}^{b}f(x){dx}=(F(b)+c)-(F(a)+c)=F(b)-F(a)$, where $F(x)$ is the indefinite integral.
So $\int_{0}^{x}{(4-t)}{dt}=|_{0}^{x}4t-\frac{1}{2}t^2=\left(4(x)-\frac{1}{2}(x)^2+c\right)-(4(0)-\frac{1}{2}(0)+c)=-\frac{1}{2}x^2+4x$
Now just substitute the values $x=0,1,2,3$
Finally for $A'(x)$ just take the derivative of the resulting function.
$\frac{d}{dx}\left(-\frac{1}{2}x^2+4x\right)=-x+4$
Then substitute $x=0,1,2,3$ once again.
