# What is physical interpretation gives integration?

It is my understanding that the integration is the inverse process of differentiation and its meaning is a fine sum (in fact, so is its symbol) but what physical interpretation do we get from this? At least, speaking of position, velocity and acceleration.

• en.wikipedia.org/wiki/Antiderivative
– user65081
Jan 2, 2015 at 14:49
• Ummmm... The Fundamental Theorem of the Calculus? Jan 2, 2015 at 17:46
• @Wolphramjonny : no Jonny, let's not complicate the explanation with antiderivative and other things. I am trying here to keep simple. So I think that it's better for the moment. The OP asks for intuitive meanings. Jan 3, 2015 at 12:07

So, you ask "if the derivative tells me how to change a variable relative to each other (e.g. position in time) what about the integral? physically!"

For intuitiveness I will refer to the function $v(t)$, the velocity of a car. For simplicity, let's assume that the road is straight. Then we have $dv(t)/dt = a(t)$, the acceleration. However, the integral

(1) $\frac {1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) dt$

will give us the distance travelled by the car between the times $t_1$ and $t_2$, divided by $t_2 - t_1$, i.e. the mean velocity between these times.

So, integration doesn't serve only for calculating areas, volumes, etc. it is also used for averaging.

Now, you will maybe want to know how do get along the average in (1) and the velocity itself. Then assume that $t_1$ and $t_2$ are very close, i.e. $t_2 = t_1 + dt$ where $dt$ is very small. Then (1) becomes

(2) $\frac {1}{dt} \int_{t_1}^{t_1 +dt} v(t) dt = \frac {1}{dt} v(t) dt = v(t)$

If the domain over which we average is very small, we get the function itself.

• This does not look very different from your other answer, so why is it a 2nd answer? Why not just add equation (2) to your first answer? Note also that $$\int v(t)\,dt\tag{1}$$ will give you a right-adjusted label. Jan 3, 2015 at 15:03
• @KyleKanos : no, first of all his text of the question was hardly understandable. I answered it. But he asked for some clarifications that look as a new question. The lack of clarity of the language in his question leaves much space to interpretation of what he asked. Jan 3, 2015 at 15:17
• @KyleKanos: since you are here, can you tell me how to mark a matrix? I didn't know, I wanted to edit some matrix and whatever I could do was to edit each line of it separately. But, how to edit the undulated parentheses between which all the matrix is enclosed? I had the same problem with marking a column vector. How to put a parenthesis over three rows? Jan 3, 2015 at 15:21
• For Latex/MathJax matrices: $$\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$$. That will get you a 2x2. For higher dimensions, use the same number of c's as dimensions, & separates the columns and \ gives a new row. Jan 3, 2015 at 15:26

Integration has different meanings. First of all is was developed for the need of calculating surface, see the achievements in the ancient world, Greece, China, http://en.wikipedia.org/wiki/Integral#Pre-calculus_integration.

So, if you draw a curve in the $x, y$ plane, then $\int_a^b f(x) dx$ is the area under the curve between the verticals $x = a$ and $x = b$, see picture.

But the integral $\frac {1}{|b-a|}\int_a^b f(x) dx$ has also another meaning: mediation, i.e. the mean value of $f(x)$ in the interval $[a, b]$. Calculi of mean values of different quantities are done this way, for instance if you want to know the mean value of a quantity X with continuous values, given that the value X=x appears with probability $p(x) dx$, then $<X> = \int x p(x) dx$, where the integral is carried over all the values that X can take.

On the other hand, the derivative $y = df/dx|_{x = c}$ has the meaning of the direction of the tangent to the curve $y=f(x)$ in the point $x = c$. If, for instance, you are given the position of a car on a road as a function of time, $r = f(t)$, then its velocity at some time $t_1$ is $v = [df/dt]_{t = t_1}$.

• okay, this is how important mathematics of the integral, its relationship with the derivative and the area between curves, but not only speak of this integral, also spoke of the vector integral multiple integrals, and to be more specific, if the derivative tells me how to change a variable relative to each other (eg position in time) what about the integral? physically! Jan 3, 2015 at 3:29
• @CristianRodríguez : it's a bit difficult to understand what you ask, because of the English. I can only guess. The generalization from single integral to multiple integral is immediate. Integrating the space beneath a surface you get the volume. Imagine that my upper figure is not in two dimensions, $x$ and $y$, but in three, $x, y, z$, and instead of the curve $y = f(x)$, you have a surface, $z = f(x, y)$. Then, the integral $\int f(x, y)\ dx \ dy$ gives you the volume under that surface. Jan 3, 2015 at 11:51
• @CristianRodríguez (continuation) In physics we need sometimes to mediate a function like $f(x_1, x_2, ... x_n)$ over a generalized volume $V$ in the $n$-dimensional space occupied by the variables $x_1, x_2, ... x_n$. If so, the mean value is $(1/V) \ \int dx_1 \int dx_2 ...\int dx_n \ f(x_1, x_2, ... x_n)$. Jan 3, 2015 at 11:59
• @CristianRodríguez You also ask if the derivative tells us how changes a variable with respect to another one? Yes, it's a bit complicated but we can do this. The space of a comment is not sufficient to me, so please see my additional answer. Jan 3, 2015 at 12:02
• Excuse my English, I'm Hispanic and I have never taken proper English courses, I think you're the person who has approached what I look mean ... Thanks to all by striving to understand and explain to me. Jan 3, 2015 at 15:11

An intuitive understanding of integration can be obtained by considering the example of two graphs(functions) : one a plot of distance($d$) covered at any given point in time (let's call this variable $t$) by an object, and on another graph we have a plot of the velocity of the object against time. You give me a time (argument to the function) and I can give you a distance and a velocity (the output $y$). However, we can see that the distance covered and velocity have a relationship. The distance covered will depend on the velocity of the object. So what would happen if instead of giving you the graph for distance and velocity, I only provided the graph (or function) for velocity? This is where integration comes into play. Let's say this object was moving at a speed of $10~\text{m/s}$ in $1~\text{second}$ the object will have covered $10~\text{meters}$, at $2~\text{seconds}$ the object covers $20~\text{meters}$, so on and so forth. This is an example of a constant slope (line). So if I were to ask you what the distance covered by this object at time ($t$), from time = $0$. What you would do is add the velocity of the object to that given time. Lets say I asked what the distance covered was at $t = 5~\text{seconds}$. It is logical that in order to find the distance you add the velocity over time. Which at a velocity of $10~\text{m/s}$ in $5~\text{seconds}$ the object would cover a distance of $50~\text{meters}$. However, as you have probably seen, graphs (functions) are rarely linear. An object will seldom have a constant speed. In the objects trajectory it will probably speed up and slow down. So if I asked for the distance the concept of integration still holds however instead of now taking the object's constant velocity and multiplying by how much time has progressed, you must add up all the little pieces or changes in the object's velocity in order to obtain the distance covered in a given time.

• Can you break up the wall of text a little? Jan 2, 2015 at 19:03