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Question Number 212626 by Ghisom last updated on 19/Oct/24
let f(x)=(1/( (√((x−a)(x−b)(x−c)))))  let a, b, c ∈R ∧a<b<c  ⇒ D(f(x))=(a, b)∪(c, ∞)  prove ∫_a ^b f(x)dx=∫_c ^∞ f(x)dx
letf(x)=1(xa)(xb)(xc)leta,b,cRa<b<cD(f(x))=(a,b)(c,)provebaf(x)dx=cf(x)dx
Answered by MrGaster last updated on 02/Nov/24
∫_a ^b f(x)dx=∫_a ^b (1/( (√((x−a)(x−b)(x−c)))))dx  ∫_c ^∞ f(x)dx=∫_c ^∞ (1/( (√((x−a)(x−b)(x−c)))))dxx=  a+(b−a)tdx=(b−a)dt  ∫_a ^b (1/( (√((x−a(x−b(x−c)))))dx=∫_0 ^1 (((b−a))/( (√((b−a)(b−a)t(b−a)(1−t)(b−a)(1−((a+(b−a)t−c)/(b−a)))))))dt  =∫_0 ^1 (1/( (√(t(1−t)(1−((a−c)/(b−a))−t)))))dtx=c+(x−c)t′dx=(x−c)dt′  ∫_c ^∞ (1/( (√((x−a)(x−b)(x−c)))))dx=∫_0 ^∞ (((x−c))/( (√((x−c)(x−c)t′(x−c)(1−t′)(x−c)))))dt′     =∫_0 ^∞ (1/( (√(t′(1−t′)(1+((x−c)/(c−a))−t′)))))dt′  t′=1−tdt′=−dt  ∫_0 ^∞ (1/( (√(t′(1−t′)(1+((x−c)/(c−a))−t′)))))dt′=∫_0 ^(−∞) ((−1)/( (√((1−t)(t)(1+((x−c)/(c−a))−1+t)))))dt  =∫_(−∞) ^1 (1/( (√((1−t)(t)(1+((x−c)/(c−a))−1+t)))))dt=∫_(−∞) ^1 (1/( (√((1−t)(t)(t+((x−c)/(c−a)))))))dt=∫_(−∞) ^1 (1/( (√((1−t)(t)(t+((b−c)/(b−a)))))))dt=∫_0 ^1 (1/( (√(t(1−t)(1−((a−c)/(b−a)))))))dt  ∫_a ^b f(x)dx=∫_c ^∞ f(x)dx
abf(x)dx=ab1(xa)(xb)(xc)dxcf(x)dx=c1(xa)(xb)(xc)dxx=a+(ba)tdx=(ba)dtab1(xa(xb(xc)dx=01(ba)(ba)(ba)t(ba)(1t)(ba)(1a+(ba)tcba)dt=011t(1t)(1acbat)dtx=c+(xc)tdx=(xc)dtc1(xa)(xb)(xc)dx=0(xc)(xc)(xc)t(xc)(1t)(xc)dt=01t(1t)(1+xccat)dtt=1tdt=dt01t(1t)(1+xccat)dt=01(1t)(t)(1+xcca1+t)dt=11(1t)(t)(1+xcca1+t)dt=11(1t)(t)(t+xcca)dt=11(1t)(t)(t+bcba)dt=011t(1t)(1acba)dtabf(x)dx=cf(x)dx
Commented by Ghisom last updated on 02/Nov/24
you write  a+(b−a)tdx=(b−a)dt  but this makes no sense. please correct  and clarify your answer.
youwritea+(ba)tdx=(ba)dtbutthismakesnosense.pleasecorrectandclarifyyouranswer.

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