Definite integral of a function f(x) over the interval [a, b] is written as:
$$\int_{a}^{b} f(x)dx.$$
If function y = f(x) is continuous real-valued function defined on a closed interval [a; b], then this function is differentiable on the open interval (a, b), and:
$${F}'(x)=\frac{d}{dx}\int_{a}^{x} f(t)dt=f(x).$$
If function f(x) is Riemann integrable on [a, b] and exists antiderivative F(x) of f(x) in [a, b], then:
$$\int_{a}^{b} f(x)dx=F(b)-F(a).$$
$$\int_{a}^{b} 1dx=b-a$$
$$\int_{a}^{b} c\times f(x)dx=c\times \int_{a}^{b} f(x)dx$$
$$\int_{a}^{b} [f(x) \pm g(x)]dx=\int_{a}^{b} f(x)dx \pm \int_{a}^{b} g(x)dx$$
$$\int_{a}^{b} f(x)dx=-\int_{b}^{a} f(x)dx$$
$$\int_{a}^{a} f(x)dx=0$$
For each of the three numbers a, b, and c, in the region where the function f (x) is integrable, the equation holds:
$$\int_{a}^{b} f(x)dx=\int_{a}^{c} f(x)dx+\int_{c}^{b} f(x)dx$$
$$\int_{a}^{b} [\alpha f(x) + \beta g(x)]dx=\alpha \int_{a}^{b} f(x)dx + \beta \int_{a}^{b} g(x)dx$$
If f(x) ≤ g(x) for each x ∈ (a, b), then:
$$\int_{a}^{b} f(x)dx\leq \int_{a}^{b} g(x)dx$$