# Nabla integral

Theorem: The following formula holds: $$\int_{\rho(t)}^{t} f(\tau) \nabla \tau = \nu(t)f(t)$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t)+g(t) \nabla t = \int_a^b f(t) \nabla t + \int_a^b g(t) \nabla t$$

Proof:

Theorem: The following formula holds: $$\int_a^b \alpha f(t) \nabla t = \alpha \int_a^b f(t) \nabla t$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t) \nabla t = -\int_b^a f(t) \nabla t$$

Proof:

Theorem: The following formula holds: $$\int_a^b f(t)\nabla t = \int_a^c f(t) \nabla t +\int_c^b f(t) \nabla t$$

Proof:

Theorem (Integration by parts,I): The following formula holds: $$\int_a^b f(t)g^{\nabla}(t) \nabla t = (fg)(b)-(fg)(a)-\int_a^b f^{\nabla}(t)g(\rho(t)) \nabla t.$$

Proof:

Theorem (Integration by parts,II): The following formula holds: $$\int_a^b f(\rho(t))g^{\nabla}(t) \nabla t = (fg)(b)-(fg)(a)-\int_a^b f^{\nabla}(t)g(t) \nabla t.$$

Proof:

Theorem: The following formula holds: $$\int_a^a f(t) \nabla t = 0$$

Proof:

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

Proof:

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

Proof: