Why does an integral change signs when flipping the boundaries?

Here's another intuitive justification. The obvious graphical intuition says that when $a \leq b \leq c$, then $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$. If we want this formula to hold for arbitrary $a,b,c$, then we should be able to take $a=c$, so that $\int_a^b f(x) dx + \int_b^a f(x) dx = \int_a^a f(x) dx$. But $\int_a^a f(x) dx = 0$, so if we want this formula to hold, we need $\int_a^b f(x) dx = -\int_b^a f(x) dx $.

$$\int_{a}^{b} f(x)dx=F(b)-F(a)=-(F(a)-F(b))=-\int_{b}^{a}f(x)dx$$ by the fundamental theorem of calculus.

Or graphically, $$-\int_{b}^{a}f(x)dx=\int_{b}^{a}-f(x)dx $$ and $$-f(x)$$ has the same area as $$f(x)$$ but under the x axis, so the signed area changes

We want $$\int_a^b +\int_b^c =\int_a^c.$$ now take $c=a$