Difference between revisions of "Math 22 Integration by Parts and Present Value"

Revision as of 07:32, 17 August 2020

 Let  $u$  and  $v$  be differentiable functions of  $x$ .
 
  $\int udv=uv-\int vdu$

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Revision as of 07:32, 17 August 2020 (view source) Tphan046 (talk \| contribs) (→‎Integration by Parts) ← Older edit		Revision as of 07:32, 17 August 2020 (view source) Tphan046 (talk \| contribs) (→‎Integration by Parts) Newer edit →
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	Let <math>u</math> and <math>v</math> be differentiable functions of <math>x</math>.		Let <math>u</math> and <math>v</math> be differentiable functions of <math>x</math>.

−	<math>\int u \dv=uv-\int v \du</math>	+	<math>\int u dv=uv-\int v du</math>