# How to think about derivatives in an abstract fashion?

Derivatives seem easy to understand abstractly as the rate of change of something, higher order derivatives are the rate of change of the rate of change of something, and so on.

I, however, have trouble understanding what an integral is in a general sense. It can be thought of as the sum of the infinitely small rectangles in an shape, but what is it, with respect to the initial function? I don't really understand why integration is the inverse operation of derivation. either.

-
"It can be thought of as the sum of the infinitely small rectangles in an shape" - precisely. You haven't been introduced to the notion of "area under a curve"? – J. M. Dec 8 '11 at 16:32
My analysis lecturer used to tell us that we should have been "running out of the lecture theatre screaming" when he proved that integration is the opposite of differentiation. It is, in many ways, a surprising result, and the proof isn't very basic (this was in a senior honours analysis course). – user1729 Dec 8 '11 at 16:52
The fact that "integration is the inverse operation of derivation" is (i) not quite accurate, but let that pass; and (ii) actually should be somewhat surprising. It's an unexpected connection when you think about the problems that give rise to the concepts (the 'instantaneous velocity' problem for derivatives, and the 'area problem' for integrals). – Arturo Magidin Dec 8 '11 at 16:53

The usual intuitive understanding of integration is the area under a function, or more generally, an accumulation of something. For example, here's a graphical representation of an integral:

$$s = \int_a^b f(x) dx$$

It's very important to understand why the integral is some sort of 'opposite' of the derivative (this is, of course, the fundamental theorem of calculus). Try this: as you said, a derivative is a rate of change. So if we consider the graphical interpretation of the integral shown above, what is the rate of change of the area? Well, the higher the function is, the faster the area under it increases, so the rate of change is just the value of the function. In terms of calculus, this means the derivative of the integral is the original function.

Conversely, what is the integral of the derivative? The derivative is the rate of change, and the integral is an accumulation. So if we accumulate, or collect, all of the tiny changes as the function goes along; well, that's also just the original function.

I hope these explanations help, but keep in mind that as you do more work with these concepts and learn more results that make use of them, you'll develop your own intuitive understanding of what they mean. And that understanding will be much more valuable than anything someone else can explain to you.

-

Integration can be seen as a mean value : if you have a function $f : [0,1] \rightarrow \mathbb{R}$, its mean value is $\int_0^1 f(x) dx$. That idea can be generalized to integrals over complicated domains, or for different measures (if you know a bit about measure theory).

From a probability point of view, this would translate as : the expectancy of a random variable $X$ is the integral $\int_{\Omega} X(\omega) d\mu(\omega)$, where $\Omega$ is the set of all possible events and $\mu(\omega)$ is the probability of the event $\omega$. For example, if you take a random number $\omega$ between 0 and 1, the expectancy of sin($\omega$) is $\int_0^1 sin(\omega) d\omega$.

Another interpretation is with velocity : if you move along a line at a non constant speed $v(t)$, then the distance achieved between times $t_0$ and $t_1$ is $\int_{t_0}^{t_1} v(t) dt$. (the mean speed is $\frac{\int_{t_0}^{t_1} v(t) dt}{t_1 - t_0}$ ).

If you want the rough idea of the link between integration (or more precisely, primitives) and derivation : if you draw thin rectangles, delimited by $f(a+k h)$ and $f(a+ (k+1) h)$ and sum it (essentially the usual method), and call the result $F$, then $F(a+h) - F(a) = h f(a)$ so that $\frac{F(a+h)-F(a)}{h}=f(a)$. $F$ is roughly a primitive of $f$, and its "rough" derivation is $f$.

-

To expand on @smackcrane's answer: Think of a bucket that is filling with water from a faucet. We can can think of the volume of water in the bucket as $V(t)$ measured in gallons. We can think of the water flowing out of the faucet as $f(t)$ measured in gallons/second (note that the flow can change with time depending on how much we open and close the valve). So by intuition the rate of change of the volume is equivalent to the flow out of the faucet. So

$$f(t) = \frac{d}{dt}V(t).$$

Make sure you pay attention to the units.

Conversely if we want to know how much volume is added to the bucket in one minute, we need to accumulate the flow over one minute. That is for each tiny slice of time of duration $dt$ we accumulate $f(t)dt$ volume of water. If we accumulate these tiny amounts of water over a minute we get the new volume. So clearly

$$V(1) = V(0) + \int_0^1{f(t)dt}.$$

Again, look at the units.

To sum up, the derivative tells us how a quantity is changing at an instant of time and integration tells us how those changes accumulate over a finite duration to yield a new quantity. Derivatives give us rates from amounts. Integrals give us amounts from rates.

-

A very practical way of thinking of integrals and derivatives is to think of them as a way of changing the units affixed to measured quantities:

if $x(t)$ measures distance in meters, then $x'(t)$ measures velocity in $\frac{m}{s}$ and $x''(t)$ measures acceleration in $\frac{m}{s^2}$.

So we see that when we compute $\displaystyle\int \left( x''(t) \frac{m}{s^2} \right) dt$ we get $x'(t) \frac{m}{s}$: we have multiplied our previous unit of acceleration, $\frac{m}{s^2}$, by the unit of time, $s$, to get the unit of velocity, $\frac{m}{s}$.

You can now imagine what happens if you introduce a new variable $T$ related to $t$ but this $T$ measures minutes instead of seconds. So $T=1$ corresponds to $t=60$, i.e. $T(t)=\frac{t}{60}$ and $t(T) = 60T$. Here change of variables/u-substitution is useful to determine how you change units in these calculations.

-