Interchange of limits of definite integral I could not understand one of the most magical property in Definite integral which is
$$\int_{a}^{b} f(x)dx=-\int_{b}^{a} f(x)dx$$
My teacher has proved it in two ways:
Proof $1$. 
By FTC $$\int_{b}^{a} f(x)dx=F(a)-F(b)=-\left(F(b)-F(a)\right)=-\int_{a}^{b}f(x)dx$$
Proof $2$.
if $a \lt b \lt c$ then
$$ \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx$$
Put $c=a$ we get  $$ \int_{a}^{b}f(x)dx+\int_{b}^{a}f(x)dx=\int_{a}^{a}f(x)dx=0$$
Hence  $$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$$
But in Proof $2$ how can we substitute $c=a$ when our assumption is $a \lt b \lt c$
If suppose $f(x)$ is above $X$ axis, then Area from $a$ to $b$ should be same as area from $b$ to $a$ right? But why negative sign should be multiplied.
For this my teacher answered that all distances measured along Negative to Positive X axis are positive and negative if we reverse the direction and he told that its a convention.   
 A: The property $\int_{b}^{a} \cdots = -\int_{a}^{b} \cdots$ you mention, along with $\int_{a}^{a} \cdots = 0$, may be viewed as definitions. Together, these definitions guarentee the cocycle condition: If $f$ is integrable on some interval $I$, and if $a$, $b$, and $c$ are points of $I$, then
$$
\int_{a}^{b} f(x)\, dx + \int_{b}^{c} f(x)\, dx + \int_{c}^{a} f(x)\, dx = 0.
$$
Intuitively, integrating over a closed loop gives the net change in $f$ at a point, namely zero.
Your teacher's arguments may be viewed as showing that the cocycle condition is compatible with:


*

*The second fundamental theorem of calculus (a.k.a., the evaluation theorem).

*The additivity property of the integral when the interval of integration is split.
Yet a third reason for the convention is:


*If you divide an interval $[a, b]$ of integration into $n$ pieces of equal length $\Delta x = \frac{1}{n}(b - a)$, then $\Delta x > 0$ if $a < b$ (i.e., you take positive steps when integrating left-to-right), and $\Delta x < 0$ if $a > b$ (i.e., you take negative steps when integrating right-to-left).

A: The definition of an integral is NOT the area under a curve (this area is the intuitive idea behind it). It is the following limit (if it exists) : https://en.wikipedia.org/wiki/Integral#Riemann_integral
You can see that in the definition, it is assumed that $a<b$. To this point, calculating the integral from $b$ to $a<b$ does not make sense, since we didn't say what it was. It is later defined as 
$\int_b^{a}f(x)dx = - \int_a^{b}f(x)dx$
This definition allows you to have plenty of nice formulas later. 
With this definition, the property is obviously true.
