# Definite integration and the $y$ axis

With definite integration you can find the area under a curve (the area between the curve and the $x$ axis).

If you have a curve $f(x)$, and integrate it to get $g(x)$, you can get the area bounded by the $x$ axis, $x=a$, $x=b$, and $f(x)$, (where $a < b$), by doing $g(b) - g(a)$.

So you are getting the area under the curve up to $x=b$, and subtracting the area up to $x=a$, to get the area between $a$ and $b$.

But where does it get the area from to $a$ and $b$?, from the y axis? Because if $a=0$ then you will be taking away $0$ as $g(0)=0$, but in many curves there is often area between the curve and the $x$ axis, to the left of the $y$ axis?

Also why is area below the $x$ axis, negative area. But area to the left of the $y$ axis not negative?

• You still here, Jonathan? – Gerry Myerson May 27 '12 at 9:29

If you integrate (I'd rather say, antidifferentiate) $f(x)$, you don't get $g(x)$, you get $g(x)+C$ for some arbitrary constant $C$. So it's wrong to think of $g(b)$ as "the area up to $x=b$."

Let's look at an example. When you antidifferentiate $\sin x$, you could get $-\cos x$, or you could get $-\cos x+42$ (or any of a number of other correct answers, but these two will do for now). So, the area up to $\pi/2$: is it $-\cos(\pi/2)$, which is $0$? or is it $42$?

Well, neither of these is a particularly good answer, which illustrates the difficulties you get into if you think of $g(b)$ as the area up to $x=b$.

Rather, you should think of $\int_a^xf(t)\,dt=g(x)-g(a)$ as the area from $a$ to $x$; it's only definite integrals that have an unambiguous interpretation as areas.

Notice that if we use a new symbol, say, $h(x)=\int_0^xf(t)\,dt$, for the area from zero to $x$ then if $f$ is always positive and $a$ is negative we'll have $h(a)$ negative, so in that sense area to the left of the $y$-axis is negative. But what this really shows is that you can't always interpret $\int_a^bf(x)\,dx$ as an area; you can always calculate the definite integral, but the number that comes out will not be what we generally accept as the area unless $a\le b$ and $f(x)\ge0$ for $a\le x\le b$.

With the given definition your calculated area will equal the area between the curve and the $x$-axis (as it is in fact stated in your definition). Now if at some point between $x=a$ and $x=b$ the curve is negative, i.e. it is "to the left of the $\mathbf x$-axis", you will have to split the integration and integrate the negative part of the curve separately then add the absolute value of the negative part to the positive part.

This is because when you calculate the area using integration bound by the $x$-axis and some curve $f(x)$ that $f(x)<0$, the result will be negative. So you must treat cases of $f(x)<0 \text{ and }f(x)>0$ separately, and always take the absolute value (since area can't be negative!).

Given an "analytical expression" $f(x)$ one can calculate (or look up in a table) its antiderivatives $F(x)+C$. These are again analytical expressions, with the property that their formal derivative (computed according to the rules) coincides with the given $f(x)$; i.e., $F'(x)=f(x)$ as expressions. There is nothing to get excited about.

Now such an $f(x)$ determines a graph $\gamma:\ x\mapsto\bigl(x, f(x)\bigr)$ in the $(x,y)$-plane. When $f(x)\geq0$ for $\,a\leq x\leq b$ it is a reasonable geometric problem to find the area$(B)$ of the set $$B:=\bigl\{(x,y)\, \bigm|\, a\leq x\leq b, \ 0\leq y\leq f(x)\bigr\}\ .$$

To solve this problem one introduces intermediary points $x_k$ $\,(0\leq k\leq N)$ such that $a=x_0<x_1<\ldots<x_N=b$ and sampling points $\xi_k\in[x_{k-1},x_k]$ $\,(1\leq k\leq N)$. Geometrical considerations then show that $${\rm area}(B)\doteq \sum_{k=1}^N f(\xi_k)(x_k-x_{k-1})\ ,$$ so that one is lead to the conclusion that area$(B)$ is in fact the limit of such sums for ever finer partitions. How could one compute such a complicated limit?

Here comes the miracle, the famous Fundamental Theorem of Calculus: The limit of these Riemann sums is equal to $F(b)-F(a)$; whence one has $${\rm area}(B)=F(b)-F(a)\ .$$ Note that the Riemann sums as well as the function $F$ make sense also when $f$ may take negative values. In this case the geometric interpretation in terms of area makes less sense. But there are also physical interpretations available, e.g., if $f(t)$ denotes the temperature at time $t$, and we want to compute the average temperature over a time interval $[a,b]$. Temperatures may well be negative.