Because the The Definite Integral is a limit of sums, it inherits the arithmetic of sums. These properties let us compute new integrals from known ones without any new limits.
Key idea
$$\int_a^a f\,dx = 0, \qquad \int_b^a f\,dx = -\int_a^b f\,dx, \qquad \int_a^b f\,dx + \int_b^c f\,dx = \int_a^c f\,dx,$$ $$\int_a^b (cf + dg)\,dx = c\int_a^b f\,dx + d\int_a^b g\,dx, \qquad m \le f \le M \text{ on } [a, b] \implies m(b - a) \le \int_a^b f\,dx \le M(b - a).$$
Why it works
Each property is a fact about sums. Reversing the limits makes every $\Delta x = \frac{b - a}{n}$ negative, flipping the sign. Splitting $[a, c]$ at $b$ splits the rectangles into two groups. Sums of $cf + dg$ separate into $c$ times one sum plus $d$ times another. And if every height is between $m$ and $M$, the total area is between the areas of the two rectangles of heights $m$ and $M$.
Worked example: Splitting and reversing
Given $\int_0^5 f\,dx = 8$ and $\int_0^2 f\,dx = 3$, find $\int_2^5 f\,dx$ and $\int_5^2 f\,dx$.
- $\int_0^2 f + \int_2^5 f = \int_0^5 f$, so $\int_2^5 f\,dx = 8 - 3 = 5$.
- Reversing: $\int_5^2 f\,dx = -5$.
Answer: $5$ and $-5$
Worked example: Linearity
With $\int_0^5 f\,dx = 8$, find $\int_0^5 (3f(x) - 2)\,dx$.
- $3\int_0^5 f\,dx - \int_0^5 2\,dx = 3(8) - 2(5)$.
- $= 24 - 10 = 14$.
Answer: $14$
Worked example: Bounds without computing
Show that $2 \le \int_0^2\sqrt{1 + x^3}\,dx \le 6$.
- On $[0, 2]$, $1 + x^3$ runs from 1 to 9, so $1 \le \sqrt{1 + x^3} \le 3$.
- Multiply by the length 2: $1 \cdot 2 \le \int \le 3 \cdot 2$.
Answer: The integral is between 2 and 6 (in fact it is about 3.24).
The bounds property gives quick sanity checks. If your computed value of $\int_0^2\sqrt{1 + x^3}\,dx$ came out as 7, the bounds above would tell you at once that something went wrong. A related property: if $f(x) \le g(x)$ on $[a, b]$, then $\int_a^b f\,dx \le \int_a^b g\,dx$, since every rectangle for $f$ is no taller than the matching one for $g$.
Common mistake
Splitting a product: $\int fg\,dx$ is not $\int f\,dx \cdot \int g\,dx$. For example, $\int_0^1 x \cdot x\,dx = \frac{1}{3}$, while $\left(\int_0^1 x\,dx\right)^2 = \frac{1}{4}$. Linearity covers sums and constant multiples only. Check with a small example before using a "property" you are unsure of.