Bacteria Problem: population starts at 7000, and growth rate in bacteria per hour is given Problem is as follows:
A bacterial population is $7000$ at time
$t = 0$,
and its rate of growth is
$1,000 · 4^t$
bacteria per hour at time $t$ hours.
What is the population after one hour? (Round your answer to the nearest whole number.)
My approach:
I'm having a hard time understanding how this involves calculus at all. It seems to me as simple as $1,000*4^(1) + 7,000 = 11,000.$ But it's obviously wrong.
Why do we need to integrate anything if we already have the rate of growth?
 A: To find the population after 1 hour, you need to find an expression for the number of bacteria as a function of $t$. What you have is the rate of growth as a function of time.
Just as an analogy, it is the same as position $s$ and velocity $v$ as a function of time. If I provide with an initial position $s_0$, and the velocity function $v(t)=\frac{ds(t)}{dt}$, and I ask you the position of an object at $t=1$, you can't say it is $s_0+v(1)$.
In order to recover how much the position of the object has changed, based on velocity, i.e. the rate of change of position $\frac{ds(t)}{dt}$, you have to do integration on the velocity function.
So, at $t=1$, the position of the object is actually $$s(1) = s_0+\int_{t=0}^{1}v(t)dt$$
Does that make sense?
A: 
Why do we need to integrate anything if we already have the rate of growth?

Because there is a Fundamental Theorem of Calculus that says that is how it works.
The number of bacteria is supposed to be some function of time.
Let $F(t)$ be the number of bacteria after $t$ hours. Note that $t$ is a real number within some interval (at least the interval $[0,1]$ since we're given the population at $t=0$ and we're asked about the population at $t=1$).
For example, after one minute the population is $F\left(\frac1{60}\right).$
The plot of the function $F(t)$ is a curved line.
If you had been given a constant rate of growth, like the slope $m$ of a line,
the plot of $F(t)$ would be a straight line and you could compute points along
it using an equation related to lines (as you seem to be trying to do),
$$ y = y_0 + m (x - x_0), $$
but you were given a rate that is different at different values of the time $t$:
the rate at one hour after the start is $4000,$ but the rate half an hour after the start is $1000 \cdot 4^{1/2} = 2000,$ which is quite a bit less;
and the rate fifteen minutes after the start is
$1000 \cdot 4^{1/4} \approx 1414,$ even less still.
So the plot of $F(t)$ is not a straight line and the formula for a straight line is useless.
You don't (initially) know what the function $F(t)$ is, but you do know its derivative:
$$ \frac{\mathrm d}{\mathrm dt} F(t) = 1000 \cdot 4^t. $$
You may also have learned the relevant part or version of the Fundamental Theorem of Calculus (some authors give you a theorem with two parts, some give you a "first theorem" and "second theorem"); possibly you learned the relevant calculation without it being called a theorem. The theorem says that if we have functions $f(t)$ and $F(t)$ with the relationship
$$ f(t) = \frac{\mathrm d}{\mathrm dt} F(t), $$
then
$$ F(b) - F(a) = \int_a^b f(t)\,\mathrm dt. $$
On the left side of this equation you are interested in $F(1)$ and you already have $F(0)$. On the right side you know $f(t)$ for all values of $t,$ and it is a function that is not hard to integrate.
