Difference between revisions of "Nabla integral"
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(Created page with "<div class="toccolours mw-collapsible mw-collapsed" style="width:800px"> <strong>Theorem:</strong> The following formula holds: $$\int_a^b f(t)+g(t) \nabla t = \int_a^b f(t) \...") |
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Revision as of 07:34, 19 October 2014
Theorem: The following formula holds: $$\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$$
Proof: █
Theorem: The following formula holds: $$\int_a^a f(t) \nabla t = 0$$
Proof: █
