Why is Integral from $a$ to $b$ equal to negative integral from $b$ to $a$ Why is the integral from $a$ to $b$ equal to negative integral from $b$ to $a$. According to my teacher this is a definition, but definitions still have some logic/reasoning behind it, so what is the logic/reasoning in this case?
 A: From the Second Fundamental Theorem of Calculus it follows that $$ \int_{a}^{b}f(x) \,\mathrm{d}x = F(b) - F(a) $$
where $F$ is a function such that $F'= f$ and $b > a$. 
If we want to extend this result to all choices of $a$ and $b$ that define a 
closed interval in $\mathbb{R}$ we should be consistent with the theorem so
$$ \int_{b}^{a}f(x) \,\mathrm{d}x = F(a) - F(b) = -\int_{a}^{b}f(x) \,\mathrm{d}x $$
Another motivation for this is that it allows for expressions like
$$F(x)= \int_{x_0}^{x}f(t) \,\mathrm{d}t $$ to be functions defined in any real 
interval (as long as the integral makes sense of course).
A: When I first learned this, I thought of it like this.  Let's say $a < b$, and you're integrating the function $f'$, which is the derivative of some function $f$.  If you're integrating from $a$ to $b$, what you're doing is taking a sum $\int$ of a bunch of small changes in $y$:
$$\int\limits_a^b f'(x)dx = \int\limits_a^b \frac{dy}{dx}dx = \int\limits_a^b dy$$
That is, you're adding an infinite number of small changes in $y = f(x)$, like $dy = f(x_2)- f(x_1)$, for $x_1 < x_2$ infinitely close together, a little bit at a time, starting at $a$, until you get to $b$.  Then
$$\int\limits_b^a dy$$
is just the reverse process. This time you start at $b$ and work your way down to $a$.  So you'd be adding up infinitely many infinitely  small changes like $dy = f(x_2) - f(x_1)$ for $x_2 < x_1$.  Clearly each small change here is the negative of one you encountered before. So 
$$\int\limits_a^b f'(x)dx = \int\limits_a^b dy =- \int\limits_b^a dy = -\int\limits_b^a \frac{dy}{dx}dx=-\int\limits_b^a f'(x)dx$$
A: It's so that the equality$$\int_a^cf(x)\,\mathrm dx=\int_a^bf(x)\,\mathrm dx+\int_b^cf(x)\,\mathrm dx$$always holds.
A: $$\int_a^b f(x)\,dx = F(x)\Big|_a^b=F(b)-F(a)=-\Big[F(a)-F(b)\Big]=-\Big[F(x)\Big]_b^a=-\int_b^a f(x)\,dx$$
