Here we show how to use the the Fundamental Theorem of Calculus to evaluate the definite integral without calculating a Riemann sum, which works as long as we can calculate the indefinite integral, or antiderivative, $\int f(t) \, dt$.

If $f(t)$ is the rate at which a function $F(t)$ is changing, then the Riemann sum estimates the total change in $F(t)$ from $a$ to $b$. As we increase $n$, this estimate improves. The limit as $n$ goes to infinity, i.e., the definite integral $\displaystyle \int_a^b f(t) \, dt$, is the actual value of the total change in $F(t)$ from $a$ to $b$.
Now, what does it mean for $f(t)$ to be the rate at which $F(t)$ is changing? It means that $F$ satisfies the differential equation $\frac{dF}{dt} =$
.
Suppose we knew what the function $F(t)$ was, and we wanted to know the total change in $F(t)$ from $t=a$ to $t=b$. We could do this just by subtracting two values of $F$:
.

Let's look at a specific example. Suppose $f(t)=3 t^{2}  1$ and we want to know the change in $F$ from $t=0$ to $t=2$. We can solve the differential equation $\frac{dF}{dt}=3 t^{2}  1$ using an indefinite integral:
$\displaystyle F(t) = \int 3 t^{2}  1 \, dt =$
To finish solving the differential equation, we need an initial condition. Let's see what happens when we pick three different initial conditions.
First, let's try $F(0)=0$. In this case, what is $C$?
So $F(t)=$
. To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is

=
.
Next, let's try $F(0)=2$. In this case, what is $C$?
So $F(t)=$
. To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is

=
.
Now, you pick an initial condition to try: $F(0)=$
. In this case, what is $C$?
So $F(t)=$
. To find the total change, we need to subtract $F(0)$ from $F(2)$. The total change is

=
.
What effect does the initial condition have on the total change?

We can evaluate a definite integral either by taking the limit of Riemann sums or by solving a pure time differential equation. This is the idea of the Fundamental Theorem of Calculus:
If $f(t)$ is a continuous function and $F(t)$ any differentiable function with $F'(t)=f(t)$, then $$\int_a^b f(t)\, dt=F(b)F(a)=F(t)\Big _a^b$$
The notation $F(t)\Big _a^b$ is just shorthand notation for $F(b)F(a)$.
In other words, if we can find an antiderivative of $f(t)$, we can evaluate the definite integral just by plugging in the limits of integration and subtracting. (We can't always find an antiderivative, but we won't worry about such cases here.)