Regla de Barrow

Una consecuencia útil del teorema fundamental es la regla de Barrow, que permite el cálculo de una integral definida utilizando una primitiva.

 Si f(x) es una función continua en [a,b] y F(x) es una primitiva de f(x), entonces ∫ a b f(x)dx =F(b)−F(a) MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qmaeaacaWGMbGaaiikaiaadIhacaGGPaGaamizaiaadIhaaSqaaiaadggaaeaacaWGIbaaniabgUIiYdGccqGH9aqpcaWGgbGaaiikaiaadkgacaGGPaGaeyOeI0IaamOraiaacIcacaWGHbGaaiykaaaa@471D@

Usando la regla de Barrow, se puede calcular una integral definida en dos pasos:
   1. Se calcula una primitiva F(x), usando un método de integración conveniente.
   2. Se evalúa en b y en a, y luego se resta: F(b) - F(a) .

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∫ 0 5 1 x 2 +4x+6 dx MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qmaeaadaWcaaqaaiaaigdaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGinaiaadIhacqGHRaWkcaaI2aaaaiaadsgacaWG4baaleaacaaIWaaabaGaaGynaaqdcqGHRiI8aaaa@4270@  La función  f(x)= 1 x 2 +4x+6 MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maalaaabaGaaGymaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI0aGaamiEaiabgUcaRiaaiAdaaaaaaa@412F@  es continua en el intervalo [0,5].
 Primero, calculamos una primitiva de f(x). Para ello usaremos ∫ 1 1+ x 2 dx=arctan⁡x+K MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaadaWcaaqaaiaaigdaaeaacaaIXaGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaaGccaWGKbGaamiEaiabg2da9iGacggacaGGYbGaai4yaiaacshacaGGHbGaaiOBaiaadIhacqGHRaWkcaWGlbaaleqabeqdcqGHRiI8aaaa@476F@ .
 Necesitamos hacer una tranformación (completando cuadrados):  x 2 +4x+6= (x+2) 2 +2=2[ ( 1 2 x+ 2 ) 2 +1 ] MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaisdacaWG4bGaey4kaSIaaGOnaiabg2da9iaacIcacaWG4bGaey4kaSIaaGOmaiaacMcadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaeyypa0JaaGOmamaadmaabaWaaeWaaeaadaWcaaqaaiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaOGaamiEaiabgUcaRmaakaaabaGaaGOmaaWcbeaaaOGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdaaiaawUfacaGLDbaaaaa@4FE1@
 Si hacemos el cambio de variable  t= 1 2 x+ 2 MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2da9maalaaabaGaaGymaaqaamaakaaabaGaaGOmaaWcbeaaaaGccaWG4bGaey4kaSYaaOaaaeaacaaIYaaaleqaaaaa@3C4F@ ,  dt= 1 2 dx MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadshacqGH9aqpdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaikdaaSqabaaaaOGaamizaiaadIhaaaa@3C68@ ,
 resulta la primitiva ∫ dx x 2 +4x+6 = ∫ dx 2[ ( 1 2 x+ 2 ) 2 +1 ] = 1 2 ∫ dt t 2 +1 MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5A6A@   =arctan⁡t=arctan⁡( 1 2 x+ 2 ) MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaciyyaiaackhacaGGJbGaaiiDaiaacggacaGGUbGaamiDaiabg2da9iGacggacaGGYbGaai4yaiaacshacaGGHbGaaiOBamaabmaabaWaaSaaaeaacaaIXaaabaWaaOaaaeaacaaIYaaaleqaaaaakiaadIhacqGHRaWkdaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaaa@4A0E@
  Entonces ∫ 0 5 1 x 2 +4x+6 dx = [ 1 2 arctan⁡( 1 2 x+ 2 ) ] 0 5 = 1 2 [ arctan⁡( 5 2 + 2 )−arctan⁡( 2 ) ] MathType@MTEF@5@5@+=feaafiart1ev1aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6C73@

Algunas funciones no tienen primitiva; en ese caso este método no se puede aplicar. También puede ocurrir que la función tenga primitiva, pero que resulte excesivamente trabajoso calcularla. En los casos en que la primitiva no existe o se necesita demasiado tiempo para calcularla, la integral definida se calcula de forma aproximada, utilizando métodos numéricos que se basan en la relación entre integral y área.