Quiz:Integration by parts: Difference between revisions
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+ We can compute an expression for the antiderivative of the [[pointwise product of functions]] <math>fg</math> based on knowledge of expressions for <math>f</math>, <math>g</math>, and their antiderivatives. | + We can compute an expression for the antiderivative of the [[pointwise product of functions]] <math>fg</math> based on knowledge of expressions for <math>f</math>, <math>g</math>, and their antiderivatives. | ||
- Suppose <math>F</math> and <math>G</math> are everywhere differentiable. Given an expression for the antiderivative for the [[pointwise product of functions]] <math>F'G</math>, we can obtain an expression for the antiderivative for the pointwise product <math>FG'</math>. | - Suppose <math>F</math> and <math>G</math> are everywhere differentiable. Given an expression for the antiderivative for the [[pointwise product of functions]] <math>\! F'G</math>, we can obtain an expression for the antiderivative for the pointwise product <math>\! FG'</math>. | ||
- If <math>F</math> is a one-to-one function, we can find an antiderivative for <math>F^{-1}</math> in terms of <math>F^{-1}</math> and an antiderivative for <math>F</math>. | - If <math>F</math> is a one-to-one function, we can find an antiderivative for <math>F^{-1}</math> in terms of <math>F^{-1}</math> and an antiderivative for <math>F</math>. | ||
Revision as of 00:13, 29 December 2011
Key observations
Equivalence of integration problems