# What does integration do?

I know that integrals are used to compute the area under a curve. Let's say I have $y = x^2$. It creates smaller rectangles and then add up the sum (assuming that rectangles are going infinitely in number and is like going to a limit).

But I recently encountered a problem in my mind. Suppose we have a function, $y = x^2$. If we integrated it, we simply get the anti derivative of it which is $x^3/3$, assuming that the area is not of concern. What is the correlation of $x^3/3$ to $x^2$? I mean, it simply likes transforms a function into another function, but I can't get a clearer picture. When we graph $x^2$ and $x^3/3$, there is no connection visually. They are simply different graphs.

-
Maybe one of the fundamental theorems of calculus could help you out: $$\frac{d}{dx} \int f(x) dx = f(x).$$ This means that the "instantaneous slope" on the graph of $y=\text{antiderivative}$ is just $f(x)$. – anon Feb 5 '12 at 10:20
@anon I think you mean $\displaystyle \frac{d}{dx}\int_a^x f(t) dt = f(x)$ – Pedro Tamaroff Feb 5 '12 at 18:59

The word "integral" is used in two completely different senses. The first, called definite integral, has a simple geometric (or physical) interpretation, the second, called indefinite integral, is accessible only to people having the notion of "derivative of a function of one variable" in their repertoire. It is true that in the one-dimensional case there is a connection between the two notions. This connection is called the fundamental theorem of calculus.

(a) The definite integral: Given some sort of "intensity" $f(x)$ at each point $x$ of some domain $B$ (an interval, a sphere, a cube in ${\mathbb R}^n$, etc.), where $f(x)$ varies with $x$, one can ask for the "total effect" an agent of this intensity could have. This total effect is the integral of $f$ over $B$ and is denoted by $$\int_B f(x){\rm d}(x)$$ (or similar). From the geometric intuition behind it this quantity is a limit of Riemann sums, viz. $$\int_B f(x){\rm d}(x)\ =\ \lim_{\ldots} \sum_k f(\xi_k)\ \mu(B_k)\ ,$$ where the $B_k$ form a disjoint partition of $B$ into very small subdomains and $\mu$ denotes the natural geometric measure (length, surface area, $n$-dimensional volume) in the situation at hand.

(b) The indefinite integral: Given a function $t\mapsto f(t)$ on some interval $I\subset{\mathbb R}$ one may ask: Is this function the derivative of some other function $F(\cdot)$? The answer is yes, and in fact there is an infinite set of such functions $F(\cdot)$, whereby any two of them differ by a constant on $I$. This set of functions is called the indefinite integral of $f$ on $I$ and is denoted by $$\int f(t)\ dt\ .$$

(c) The fundamental theorem of calculus: Given two points $a$, $b\in I$ the difference $F(b)-F(a)$ has the same value for all functions $F\in\int f(t)\ dt$ and may as well be denoted by $$\int_a^b f(t)\ dt\ .$$ Now comes the theorem (and this is the real wonder, not the fact that taking the derivative of the antiderivative of $f$ gives back $f$): When $a<b$ then $$\int_{[a,b]} f(t)\ {\rm d}t = \int_a^b f(t)\ dt\ .$$ Here on the left side we have a limit of Riemann sums, and on the right side a difference of $F$-values.

-

The key word here is instantaneous. Although the 2 graphs are "different". They are linked to each other through the "instantaneous area". What I mean is that if you take a point on your function and calculate the value of the area of the function from, say, 0 to that point. That value of area is the same as g(x). $(g(x) = \int(f(x))$

Consider:

$x =3$, the area of the graph from 0 to 3 is the same as the value of g(x) i.e. $\int(f(x))$

-

It may be helpful to divorce (somewhat) the ideas of the integral and the antiderivative in your mind. The the definite integral is simply the concept of signed area. The integral exists, even for functions with breaks, corners, and other points generally considered 'not nice' (provided there are only countably many such points.

The antiderivative, on the other hand, is a function $F(x)$ such that $F'(x) = f(x)$.

The connection between them is the First Fundamental Theorem of Calculus, namely that: $$F(x) = \int_a^x f(t)dt$$ Without the fundamental theorem, there is no connection between antiderivative and integral. While this way of thinking is perhaps a bit extreme, it provides a good way of thinking integral and antiderivative in a more rigorous way.

-

Lets have an intuitive debate of the matter, keep in mind that some of the concepts below aren't formal, and are only applicable visually to "nice" curvy functions.

It might be easier to look at this the other way around. As you can see from examining the plot posted on other answers (and some other function plots superimposed on plots of their derivative), you could see that at every point - the larger the value of the derivative is, the larger the change in the function would be at this point. The derivative somewhat "measures" the function's tendency to change at that given point.

Now, when you discuss an integral as the differences of the values of the antiderivative - keep that concept in mind. When asking yourself how much surface is accumulated under the function graph you're actually asking yourself at every point "what is the tendency of the function describing how much surface I've accumulated up to this point to change at this point", which will obviously be greater as the value of the function in this point is higher.

Does this relation remind you of anything? Of course! It's exactly the relation between a function and it's derivative! That means that the function f(x) = "how much surface was accumulated under the graph as I reached the point x in the integrad" would be exactly the anti-derivative of the integrad.

-

Let's begin by looking at something much simpler: consider the constant function $y = h$. An antiderivative (indefinite integral) of $y$ is simply $xh$. Notice that $xh = \int_0^x h dt$. By the interpretation of definite integration as giving us the area under the curve, we see that the quantity $xh$ ought to be the area of the rectangle with base length $x$ and height $h$, and this is precisely what it is. Thus the integral agrees with the standard formula from grade school.

Now let us consider the function $y=x$. We see that an antiderivative for $x$ is $x^2/2 = \int_0^x tdt$. The right hand side indicates that we are looking at the area under the curve from $0$ to $x$, and the left hand side tells us that $x^2/2$ must be the formula for the area. Here is the neat thing- the region under the graph of $y = x$ is a triangle with base $x$ and height $x$, therefore the process of integration has recovered for us the standard formula for the area of a triangle $\frac{1}{2} bh$.

Now let us turn to your example $y = x^2$. An antiderivative, as you have mentioned, is $x^3/3 = \int_0^x t^2 dt$. Now I can answer your question: The precise relationship between the antiderivative and the definite integral is that the antiderivative $x^3/3$ is the area under the curve $y = t^2$ from $t=0$ to $t=x$.

So the antiderivative is telling you the amount of "area under the curve so far." What you have essentially done is proven an area formula (just like length $\times$ width or 1/2base $\times$ height) for the region under the parabola. I suppose your math teacher could have made you memorize this formula in grade school as well, but somehow triangles and rectangles are gometrically more natural than regions under parabolas:)

More generally, let $y = f(t) \geq 0$ be any continuous function, and consider the antiderivative $F(x) = \int_a^x f(t)dt$. The function $F(x)$ is then a formula for the area under the graph of $f(t)$ from $a$ to $x$. This is one way of interpreting the content of the Fundamental Theorem of Calculus in the case that $f(x) \geq 0$, and it is a very satisfying interpretation indeed!

-