Integral de 1/sqrt^8(4-x)^2 dx
Solución
Solución detallada
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Hay varias maneras de calcular esta integral.
Método #1
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Vuelva a escribir el integrando:
(4−x)641=x32−128x31+7936x30−317440x29+9205760x28−206209024x27+3711762432x26−55146184704x25+689327308800x24−7352824627200x23+67645986570240x22−541167892561920x21+3788175247933440x20−23311847679590400x19+126550030260633600x18−607440145251041280x17+2581620617316925440x16−9719042324016660480x15+32396807746722201600x14−95485328095602278400x13+248261853048565923840x12−567455664111007825920x11+1134911328222015651840x10−1973758831690462003200x9+2960638247535693004800x8−3789616956845687046144x7+4081125953526124511232x6−3627667514245444009984x5+2591191081603888578560x4−1429622665712490250240x3+571849066284996100096x2−147573952589676412928x+184467440737095516161
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Vuelva a escribir el integrando:
x32−128x31+7936x30−317440x29+9205760x28−206209024x27+3711762432x26−55146184704x25+689327308800x24−7352824627200x23+67645986570240x22−541167892561920x21+3788175247933440x20−23311847679590400x19+126550030260633600x18−607440145251041280x17+2581620617316925440x16−9719042324016660480x15+32396807746722201600x14−95485328095602278400x13+248261853048565923840x12−567455664111007825920x11+1134911328222015651840x10−1973758831690462003200x9+2960638247535693004800x8−3789616956845687046144x7+4081125953526124511232x6−3627667514245444009984x5+2591191081603888578560x4−1429622665712490250240x3+571849066284996100096x2−147573952589676412928x+184467440737095516161=(x−4)321
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que u=x−4.
Luego que du=dx y ponemos du:
∫u321du
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Integral un es n+1un+1 when n=−1:
∫u321du=−31u311
Si ahora sustituir u más en:
−31(x−4)311
Método #2
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Vuelva a escribir el integrando:
(4−x)641=x32−128x31+7936x30−317440x29+9205760x28−206209024x27+3711762432x26−55146184704x25+689327308800x24−7352824627200x23+67645986570240x22−541167892561920x21+3788175247933440x20−23311847679590400x19+126550030260633600x18−607440145251041280x17+2581620617316925440x16−9719042324016660480x15+32396807746722201600x14−95485328095602278400x13+248261853048565923840x12−567455664111007825920x11+1134911328222015651840x10−1973758831690462003200x9+2960638247535693004800x8−3789616956845687046144x7+4081125953526124511232x6−3627667514245444009984x5+2591191081603888578560x4−1429622665712490250240x3+571849066284996100096x2−147573952589676412928x+184467440737095516161
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Vuelva a escribir el integrando:
x32−128x31+7936x30−317440x29+9205760x28−206209024x27+3711762432x26−55146184704x25+689327308800x24−7352824627200x23+67645986570240x22−541167892561920x21+3788175247933440x20−23311847679590400x19+126550030260633600x18−607440145251041280x17+2581620617316925440x16−9719042324016660480x15+32396807746722201600x14−95485328095602278400x13+248261853048565923840x12−567455664111007825920x11+1134911328222015651840x10−1973758831690462003200x9+2960638247535693004800x8−3789616956845687046144x7+4081125953526124511232x6−3627667514245444009984x5+2591191081603888578560x4−1429622665712490250240x3+571849066284996100096x2−147573952589676412928x+184467440737095516161=(x−4)321
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que u=x−4.
Luego que du=dx y ponemos du:
∫u321du
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Integral un es n+1un+1 when n=−1:
∫u321du=−31u311
Si ahora sustituir u más en:
−31(x−4)311
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Añadimos la constante de integración:
−31(x−4)311+constant
Respuesta:
−31(x−4)311+constant
Respuesta (Indefinida)
[src]
/
|
| 1 1
| ----------- dx = C - -------------
| 64 31
| _______ 31*(-4 + x)
| \/ 4 - x
|
/
∫(4−x)641dx=C−31(x−4)311
Gráfica
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.