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

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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> |

## Revision as of 08:32, 17 August 2020

## Integration by Parts

Let and be differentiable functions of .Failed to parse (unknown function "\dv"): {\displaystyle \int u \dv=uv-\int v \du}

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