What splits, and what does not AB & BC

These follow from the definition as a limit of Riemann sums, and none of them requires knowing what $f$ is.

1. Zero width. $\displaystyle\int_{a}^{a} f(x)\,dx = 0$. No interval, no area, whatever $f(a)$ happens to be.
2. Reversal. $\displaystyle\int_{b}^{a} f(x)\,dx = -\int_{a}^{b} f(x)\,dx$. Running the interval backwards makes every $\Delta x$ negative, so the whole sum changes sign.
3. Additivity. $\displaystyle\int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx$.

Reversal is the one that gets lost, because on a picture the region does not move. It is not the region that changed; it is the direction of travel, and the sign is what records it. If $\int_{2}^{5} f = 9$, then $\int_{5}^{2} f = -9$, and no hypothesis about $f$ can make those two agree unless both are zero.

Additivity holds for any $c$, not only for one between $a$ and $b$, provided $f$ is integrable across everything involved. That is more useful than it looks: from $\int_{1}^{5} f = 12$ and $\int_{1}^{8} f = 20$ you get $\int_{5}^{8} f = 20 - 12 = 8$ by subtracting the shorter run from the longer one. What you may not do is add two pieces that overlap, or that leave a gap, and call the total the integral across the span.

Two more properties, and these are about the integrand rather than the interval:

$$\int_{a}^{b} k\,f(x)\,dx = k\int_{a}^{b} f(x)\,dx, \qquad \int_{a}^{b}\left[f(x) \pm g(x)\right]dx = \int_{a}^{b} f(x)\,dx \pm \int_{a}^{b} g(x)\,dx.$$

Together these are linearity, and they are what make table questions work. Given $\int_{1}^{4} f = 6$ and $\int_{1}^{4} g = -2$,

$$\int_{1}^{4}\left[3f(x) - 2g(x)\right]dx = 3(6) - 2(-2) = 22.$$

The word constant in "constant multiple" is doing real work. $k$ has to be a number, not an expression containing the variable of integration. $\int_{0}^{2} x\,f(x)\,dx$ cannot be turned into $x\int_{0}^{2} f(x)\,dx$: the left side is a number and the right side has an $x$ in it, which already settles it.

Two more comparison rules, occasionally asked directly and useful as a check:

1. If $f(x) \le g(x)$ on $[a, b]$, then $\int_{a}^{b} f \le \int_{a}^{b} g$.
2. If $m \le f(x) \le M$ on $[a, b]$, then $m(b-a) \le \int_{a}^{b} f \le M(b-a)$. Two rectangles bracket the answer.

There is no property saying $\int f g = \left(\int f\right)\left(\int g\right)$, and none saying $\int \frac{f}{g} = \frac{\int f}{\int g}$. This is not a rule with a missing hypothesis. It is not a rule.

One counterexample settles it. On $[0, 2]$,

$$\int_{0}^{2} x\,dx = 2, \qquad \int_{0}^{2} x\,dx = 2, \qquad \text{but} \qquad \int_{0}^{2} x \cdot x\,dx = \frac{8}{3} \approx 2.667,$$

and $2 \times 2 = 4$. The two numbers are not close and no adjustment brings them together.

The reason is visible in the Riemann sum. Integration adds up heights times widths, and adding is what distributes over adding. Multiplying two sums together produces every cross term, and nothing in the setup produces those.

The same objection kills two neighbours worth naming, because they look different and are the same mistake:

1. Composites. $\int \sin(x^{2})\,dx$ is not $\int \sin x\,dx$ composed with anything. A composite is not a product, and it does not come apart either.
2. Powers. $\int \left[f(x)\right]^{2}dx$ is not $\left[\int f(x)\,dx\right]^{2}$, since a square is a product of a thing with itself.

Products can be integrated; they just need a technique rather than a property. Substitution in Topic 6.9 handles the ones where one factor is the derivative of the inside of the other, and integration by parts in Topic 6.11 handles a wide class of the rest.

On an interval $[-a, a]$, symmetry of the integrand collapses the work.

1. If $f$ is even, meaning $f(-x) = f(x)$, then $\displaystyle\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$. The two halves are mirror images and both count positively.
2. If $f$ is odd, meaning $f(-x) = -f(x)$, then $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$. The halves are mirror images through the origin, so one is the negative of the other.

Both need the interval centred at $0$. $\int_{-1}^{3} x^{3}\,dx$ is not zero, because $[-1, 3]$ is not symmetric, and there is no reason for the pieces to cancel.

Linearity lets you use symmetry on part of an integrand. For $\int_{-3}^{3}\left(x^{3} + x^{2}\right)dx$, split it: the $x^{3}$ term is odd and contributes $0$, and the $x^{2}$ term is even and contributes $2\int_{0}^{3} x^{2}\,dx = 2(9) = 18$. Total $18$, with one of the two integrals never computed.

The odd case comes with the same warning Topic 6.1 gave. A zero integral means the signed areas cancelled, not that the region is empty. If a question asks for the total area between $y = x^{3}$ and the axis on $[-3, 3]$, the answer is $2\left(\frac{81}{4}\right) = 40.5$, and the integral being zero is exactly why you have to compute it in halves.