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?
On
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.
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?