Difference between revisions of "Nabla integral"
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<strong>Theorem:</strong> The following formula holds: | <strong>Theorem:</strong> The following formula holds: | ||
$$\int_a^a f(t) \nabla t = 0$$ | $$\int_a^a f(t) \nabla t = 0$$ | ||
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<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
<strong>Proof:</strong> █ | <strong>Proof:</strong> █ | ||
Latest revision as of 02:08, 5 May 2015
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: █
