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?
Best Answer
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 $a\le x\le b$.