Delta integral

There are a few equivalent definitions of $\Delta$-integration.

Cauchy $\Delta$-integral

Let $\mathbb{T}$ be a time scale. We say that $f$ is regulated if its right-sided limits exist (i.e. are finite) at all right-dense points of $\mathbb{T}$ and its left-sided limits exist (i.e. are finite) at all left-dense points of $\mathbb{T}$. We say that $f$ is pre-differentiable with region of differentiation $D$ if $D \subset \mathbb{T}^{\kappa}$, $\mathbb{T}^{\kappa} \setminus D$ is countable with no right-scattered elements of $\mathbb{T}$, and $f$ is $\Delta$-differentiable at each $t \in D$. Now suppose that $f$ is regulated. It is known that there exists a function $F$ which is pre-differentiable with region of differentiation $D$ such that $F^{\Delta}(t)=f(t)$. We define the indefinite integral of a regulated function $f$ by $$\displaystyle\int f(t) \Delta t = F(t)+C$$ for an arbitrary constant $C$.

Now we define the definite integral, i.e. the Cauchy integral, by the formula $$\displaystyle\int_s^t f(\tau) \Delta \tau = F(t)-F(s)$$ for all $s,t \in \mathbb{T}$.

A function $F \colon \mathbb{T}\rightarrow \mathbb{R}$ is called an antiderivative of $f \colon \mathbb{T}\rightarrow \mathbb{R}$ if $F^{\Delta}(t)=f(t)$ for all $t \in \mathbb{T}^{\kappa}$. It is known that all rd-continuous functions possess an antiderivative, in particular if $t_0 \in \mathbb{T}$ then $F$ defined by $$F(t) = \displaystyle\int_{t_0}^t f(\tau) \Delta \tau$$ is an antiderivative of $f$.

Properties of $\Delta$-integrals

Delta integral from t to sigma(t)
Delta integral is linear
Interchanging limits of delta integral
Delta integrals are additive over intervals
Integration by parts for delta integrals with sigma in integrand
Integration by parts for delta integrals with no sigma in integrand
Delta integral over degenerate interval

Theorem: If $|f(t)| \leq g(t)$ on $[a,b)$ then $$\left| \int_a^b f(t) \Delta t \right| \leq \int_a^b g(t) \Delta t$$

Proof: █

Theorem: If $f(t) \geq 0$ for all $a \leq t < b$ then $$\displaystyle\int_a^b f(t) \Delta t \geq 0.$$

Proof: █

Theorem (Fundamental theorem of calculus,I): The following formula holds: $$\int_a^b f^{\Delta}(t) \Delta t = f(b)-f(a).$$

Proof: █

Theorem (Fundamental theorem of calculus,II): The following formula holds: $$\left( \int_{t_0}^x f(\tau) \Delta \tau) \right)^{\Delta} = f(x).$$

Proof: █