Quiz:Integration by parts: Difference between revisions
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{Suppose <math>p</math> is a polynomial function. In order to find the indefinite integral for a function of the form <math>x \mapsto p(x)\exp(x)</math>, the general strategy, which always works, is to take <math>p(x)</math> as the part to differentiate and <math>\exp(c)</math> as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works? | {Suppose <math>p</math> is a polynomial function. In order to find the indefinite integral for a function of the form <math>x \mapsto p(x)\exp(x)</math>, the general strategy, which always works, is to take <math>p(x)</math> as the part to differentiate and <math>\exp(c)</math> as the part to integrate, and keep repeating the process. Which of the following is the best explanation for why this strategy works? | ||
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- <math>\exp</math> can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way. | - <math>\exp</math> can be repeatedly differentiated and polynomials can be repeatedly integrated, giving polynomials all the way. | ||
Revision as of 23:40, 17 February 2012
For background, see integration by parts.
Key observations
Equivalence of integration problems
Specific integration types