228. Given the function $f(x)=2 x+4$. Find the increment of any of its antiderivatives when $x$ changes from -2 to 0.
Solution. Let's find the antiderivative of the given function:

$$
F=\int(2 x+4) d x=x^{2}+4 x+C
$$

Consider, for example, the antiderivatives $F_{1}=x^{2}+4 x, F_{2}=x^{2}+4 x+2$, $F_{3}=x^{2}+4 x-1$, and compute the increment of each of them on the interval $[-2,0]: \quad F_{1}(0)-F_{1}(-2)=0-(-4)=4 ; \quad F_{2}(0)-F_{2}(-2)=2-(-2)=4 ;$ $F_{3}(0)-F_{3}(-2)=1-(-5)=4$. Thus, the increments of these antiderivatives are equal to each other.

Now consider the increment of any antiderivative $F=x^{2}+4 x+$ $+C$ on the interval $[-2,0]$. We have

$$
\left(x^{2}+4 x+C\right)_{x=0}-\left(x^{2}+4 x+C\right)_{x=-2}=4
$$

Therefore, the increment of any antiderivative of the function $f(x)=$ $=2 x+4$ when $x$ changes from -2 to 0 is 4, i.e., the increments of all antiderivatives of this function are equal to each other and do not depend on $C$.

Let's find the increment of any antiderivative of the function $f(x)$ when the argument $x$ changes from $x=a$ to $x=b$:

$$
(F(x)+C)_{x=b}-(F(x)+C)_{x=0}=F(b)-F(a)
$$

The obtained result means that when the argument $x$ changes from $x=a$ to $x=b$, all antiderivatives of the given function have the same increment, equal to $F(b)-F(a)$.

This increment is called the definite integral.

Definition 2. If $F(x)+C$ is an antiderivative of $f(x)$, then the increment $F(b)-F(a)$ of the antiderivatives when the argument $x$ changes from $x=a$ to $x=b$ is called the definite integral and is denoted by the symbol

$$
\begin{aligned}
& \text { } \int_{a}^{b} f(x) d x, \text { i.e., } \\
& \qquad \int_{a}^{b} f(x) d x=F(b)-F(a)
\end{aligned}
$$

where $a$ is the lower limit, and $b$ is the upper limit of the definite integral.

The symbol $\int_{a}^{b} f(x) d x$ is read as: "the definite integral from $a$ to $b$ of $f$ of $x$ $d x$."

The function $f(x)$ is assumed to be continuous in the interval of the argument $x$ from $a$ to $b$.

To compute the definite integral $\int_{a}^{b} f(x) d x$, one finds:

1) the indefinite integral $\int f(x) d x=F(x)+C$;
2) the value of the integral $F(x)+C$ at $x=b, C=0$, i.e., compute $F(b)$
3) the value of the integral $F(x)+C$ at $x=a, C=0$, i.e., compute $F(a)$;
4) the difference $F(b)-F(a)$.

The process of computation is evident from the formula

$$
\int_{a}^{b} f(x) d x=\left.F(x)\right|_{a} ^{b}=F(b)-F(a)
$$

Equality (2) is called the Newton-Leibniz formula.

Remarks. 1. Under $F(x)$ in formula (2), the simplest of the antiderivatives is meant, for which $C=0$.