Quiz:Integration by parts: Difference between revisions
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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. | ||
+ <math>\exp</math> can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero. | + <math>\exp</math> can be repeatedly integrated and polynomials can be repeatedly differentiated, eventually becoming zero. | ||
| Each application of integration by parts results in the polynomial getting differentiated and the exponential function getting integrated. Every time the polynomial function is differentiated, its degree goes down by one. Once it becomes constant, it becomes zero in the step after that. | |||
- <math>\exp</math> and polynomials can both be repeatedly differentiated. | - <math>\exp</math> and polynomials can both be repeatedly differentiated. | ||
- <math>\exp</math> and polynomials can both be repeatedly integrated. | - <math>\exp</math> and polynomials can both be repeatedly integrated. | ||
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- <math>a + b</math> is an integer. | - <math>a + b</math> is an integer. | ||
+ <math>a - b</math> is an integer. | + <math>a - b</math> is an integer. | ||
|| For simplicity, assume <math>a < b</math> (the process works exactly the same way in reverse if <matH>b < a</math>). Start with the integral <math>\int x^be^x \, dx</math>. Now apply integration by parts taking <math>e^x</math> as the part to integrate and <math>x^b</math> as the part to differentiate. After one application of integration by parts, we need to integrate <math>x^{b-1}e^x</math>. Proceed in the way and we see that we get the integrations of <math>x^be^x, x^{b-1}e^x, x^{b-2}e^x, \dots</math>. If <math>a,b</math> differ by an integer, then after finitely many steps, we will land up with <math>\int x^a e^x\, dx</math>. | |||
- <math>ab</math> is an integer. | - <math>ab</math> is an integer. | ||
- <math>a/b</math> is an integer. | - <math>a/b</math> is an integer. | ||
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- <math>a - b = 1</math> | - <math>a - b = 1</math> | ||
+ <math>ab = 1</math> | + <math>ab = 1</math> | ||
|| Using integration by parts once, we can convert <math>\int x^a e^x\, dx</math> to <math>\int ax^{a-1} e^x \, dx</math>. Now, put <math>u = x^a</math>. Then <math>x = u^{1/a}</math>, and <math>du = ax^{a-1} \, dx</math>. So, we get that the integral is <math>\int e^{u^{1/a}} \, du</math>. Replace the dummy variable <math>u</math> by the dummy variable <math>x</math>, to obtain <math>\int e^{x^{1/a}} \, dx</math>, which is <math>\int e^{x^b} \, dx</math> by the assumption that <math>b = 1/a</math>. | |||
- <math>a/b = 1</math> | - <math>a/b = 1</math> | ||
</quiz> | </quiz> | ||
Revision as of 03:24, 20 February 2012
For background, see integration by parts.
Key observations
Equivalence of integration problems
Specific integration types