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|<math> \frac{d}{d \alpha} \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } F ( x , \alpha ) d x = \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } \frac{\partial F}{\partial \alpha} d x + F ( \Phi_2 , \alpha ) \frac{d \Phi_1}{d \alpha} - F ( \Phi_1 , \alpha ) \frac{d \Phi_2}{d \alpha}</math>
 
|<math> \frac{d}{d \alpha} \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } F ( x , \alpha ) d x = \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } \frac{\partial F}{\partial \alpha} d x + F ( \Phi_2 , \alpha ) \frac{d \Phi_1}{d \alpha} - F ( \Phi_1 , \alpha ) \frac{d \Phi_2}{d \alpha}</math>
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! style="background-color: rgb(238, 238, 238); background-image: none; background-repeat: repeat; background-attachment: scroll; background-position: 0% 0%; -moz-background-size: auto auto; -moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -moz-background-inline-policy: -moz-initial; font-size: 110%;" colspan="2" | Definite Integral containing rational and irrational expressions
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|<math>\int_{a}^{\infty} \frac {d x}{x^2 + a^2} = \frac{\pi}{2a}</math>
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|<math>\int_{0}^{\infty} \frac{x^{p-1} d x}{1 + x} = \frac{\pi}{\sin p \pi} \qquad 0 \ < \ p < \ 1</math>
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|<math>\int_{0}^{\infty} \frac{x^m d x}{x^n + a^n} = \frac{\pi a^{m+1-n}}{n \sin [ \frac{( m + 1 ) \pi}{n} ] } 0 \ < \ < \ m+1 \ < \ n</math>

Revision as of 21:44, 17 November 2010

Table of Definite Integrals
Definition of Definite Integral
$ \int_{a}^{b} f ( x ) d x = \lim_{n \to \infty} { f ( a ) \Delta x + f ( a + \Delta x ) \Delta x + f ( a + 2 \Delta x ) + \cdot \cdot \cdot + f ( a + ( n - 1 ) \Delta x ) \Delta x } $
$ \int_{a}^{b} f ( x ) d x = \int_{a}^{b} \frac{d}{dx} g ( x ) d x = g ( x ) |_{a}^{b} = g ( b ) - g ( a ) $
$ \int_{a}^{\infty} d x = \lim_{n \to \infty} \int\limits_{a}^{b} f ( x ) d x $
$ \int_{-\infty}^{\infty} f ( x ) d x = \lim_{a \to - \infty \ b \to \infty} \int\limits_{a}^{b} f ( x ) d x $
$ \int_{a}^{b} f ( x ) d x = \lim_{\epsilon \to \infty} \int\limits_{a}^{b - \epsilon} f ( x ) d x $
$ \int_{a}^{b} f ( x ) d x = \lim_{\epsilon \to \infty} \int\limits_{a + \epsilon}^{b} f ( x ) d x $
General Rules for Definite Integral
$ \int\limits_{a}^{b} { f ( x ) \pm g ( x ) \pm h ( x ) \pm \cdot \cdot \cdot } d x = \int\limits_{a}^{b} f ( x ) d x \pm \int\limits_{a}^{b} g ( x ) d x \pm \int\limits_{a}^{b} h ( x ) d x \pm \cdot \cdot \cdot $
$ \int_{a}^{b} c f ( x ) d x = c \int_{a}^{b} f ( x ) d x $
$ \int_{a}^{a} f ( x ) d x = 0 $
$ \int_{a}^{b} f ( x ) d x = - \int_{b}^{a} f ( x ) d x $
$ \int_{a}^{b} f ( x ) d x = \int_{a}^{c} f ( x ) d x + \int_{c}^{b} f ( x ) d x $
$ \int_{a}^{b} f ( x ) d x = ( b - a ) f ( c ) \qquad c \ is \ a \ number \ between \ a \ and \ b \ as \ long \ as \ f ( x ) \ is \ continous \ between \ a \ and \ b $
$ \int_{a}^{b} f ( x ) g ( x ) d x = f ( c ) \int\limits_{a}^{b} g ( x ) d x \qquad c \ is \ a \ number \ between \ a \ and \ b \ as \ long \ as \ f ( x ) \ is \ continous \ between \ a \ and \ b $
$ \ and \ g ( x ) \ge 0 $
Leibnitz rule for derivation
$ \frac{d}{d \alpha} \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } F ( x , \alpha ) d x = \int_{\Phi_1 ( \alpha )}^{\Phi_2 ( \alpha ) } \frac{\partial F}{\partial \alpha} d x + F ( \Phi_2 , \alpha ) \frac{d \Phi_1}{d \alpha} - F ( \Phi_1 , \alpha ) \frac{d \Phi_2}{d \alpha} $
Definite Integral containing rational and irrational expressions
$ \int_{a}^{\infty} \frac {d x}{x^2 + a^2} = \frac{\pi}{2a} $
$ \int_{0}^{\infty} \frac{x^{p-1} d x}{1 + x} = \frac{\pi}{\sin p \pi} \qquad 0 \ < \ p < \ 1 $
$ \int_{0}^{\infty} \frac{x^m d x}{x^n + a^n} = \frac{\pi a^{m+1-n}}{n \sin [ \frac{( m + 1 ) \pi}{n} ] } 0 \ < \ < \ m+1 \ < \ n $

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Dhruv Lamba, BSEE2010