Why does changing the bounds of integration result in the negative of the original bounds? If we were to have the integral from a to b if some function f(x), then that integral would represent the area under the curve from a to b, now based on the properties of integrals given to us in school, we are told that changing the bounds from b to a results in the negative of the original integral. However why would it result in the negative, you would think that it doesn't matter if you go from a to b or b to a as it would result in the same area under the curve. so my question is why is the integral negative when changing the bounds?
 A: It seems to me that is purely conventional. As far as I know, the main advantage of this convention is that the relation
$$
\int_a^b f(t) dt + \int_b^cf(t) dt = \int_a^c f(t)dt
$$
becomes true independently of the ordering of the values $a, b, c$, which is a great formal benefit.
Also note that in Lebesgue integration, the natural notion is that of integral of $f$ over a set, so one speaks of
$$
\int_{[a, b]}f(t) dt
$$
The convention for the bounds is not at all used in Lebesgue integral.
There are some domains such as complex analysis or differential forms where the natural notion is that of integral along an oriented path. The same sign property holds when the orientation of the path is reversed. In these cases the property can be proved from the definition of these integrals.
A: choose some $N< \min (a,b)$
define $F(x)$ as follows:
$F(x) = \int_0^x f(t) dt$
Then
$\int_a^b f(t) dt = \int_N^b f(t) dt - \int_N^a f(t) dt = F(b) - F(a)$
And 
$\int_b^a f(t) dt = F(a) - F(b) = - \int_a^b f(t) dt$
So, while we initially define the definite integral as the "area under the curve."  This is not entirely true.  The area under the curve is always positive.  But the integral is negative if $f(x) < 0$ over the interval, and it is negative, if you traverse it backward.  You might say that it is an area with an orientation.
