Integrală de la 1 la 2 din (x^2-3x+2)^n.
faravasile:
Cred ca glumești, sau ți-e greu să scrii enunțul complet și corect!
Trebuie sa arat ca (4n+2)integrala de la 1 la 2 din f^n(x)dx+n*integrala de la 1 la 2 din f^(n-1)(x)dx=0, unde f(x)=x^2-3x+2, f:[1,2]->R. Intradevar, nu e ex scris in totalitate, dar cu ajutorul integralei respective se rezolva ex. Sau nu?
Da, asa da. Ceea ce postasei tu nu era un exercitiu!
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[tex]I_n= \int\limits^1_2 {(x-1)^n(x-2)^n} \, dx =\int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}(x^2-3x+2) \, dx (1)
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![I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-1)^{n+1}]'(x-2)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}(x-1)^2} \, dx (2)](https://tex.z-dn.net/?f=I_n%3D+%5Cfrac%7B1%7D%7Bn%2B1%7D++%5Cint%5Climits%5E1_2+%7B%5B%28x-1%29%5E%7Bn%2B1%7D%5D%27%28x-2%29%5En%7D+%5C%2C+dx%3D%5C%5C%3D-+%5Cfrac%7Bn%7D%7Bn%2B1%7D+%5Cint%5Climits%5E1_2+%7B%28x-1%29%5E%7Bn-1%7D%28x-2%29%5E%7Bn-1%7D%28x-1%29%5E2%7D+%5C%2C+dx+%282%29 "I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-1)^{n+1}]'(x-2)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}(x-1)^2} \, dx (2)")
![I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-2)^{n+1}]'(x-1)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-2)^{n-1}(x-1)^{n-1}(x-2)^2} \, dx(3)](https://tex.z-dn.net/?f=I_n%3D+%5Cfrac%7B1%7D%7Bn%2B1%7D++%5Cint%5Climits%5E1_2+%7B%5B%28x-2%29%5E%7Bn%2B1%7D%5D%27%28x-1%29%5En%7D+%5C%2C+dx%3D%5C%5C%3D-+%5Cfrac%7Bn%7D%7Bn%2B1%7D+%5Cint%5Climits%5E1_2+%7B%28x-2%29%5E%7Bn-1%7D%28x-1%29%5E%7Bn-1%7D%28x-2%29%5E2%7D+%5C%2C+dx%283%29 "I_n= \frac{1}{n+1} \int\limits^1_2 {[(x-2)^{n+1}]'(x-1)^n} \, dx=\\=- \frac{n}{n+1} \int\limits^1_2 {(x-2)^{n-1}(x-1)^{n-1}(x-2)^2} \, dx(3)")
Se observa ca:[tex]2(x^2-3x+2)-(x-1)^2-(x-2)^2=-1\\ Din\ 2(1)+ \frac{n+1}{n} (2)+\frac{n+1}{n} (3) \ obtinem:\\ 2I_n+ \frac{n+1}{n}I_n+ \frac{n+1}{n}I_n= \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}\cdot (-1)} \, dx=-I_{n-1}\\ I_n \frac{4n+2}{n} =-I_{n-1}\\ (4n+2)I_n+nI_{n-1}=0[/tex]
Se observa ca:[tex]2(x^2-3x+2)-(x-1)^2-(x-2)^2=-1\\ Din\ 2(1)+ \frac{n+1}{n} (2)+\frac{n+1}{n} (3) \ obtinem:\\ 2I_n+ \frac{n+1}{n}I_n+ \frac{n+1}{n}I_n= \int\limits^1_2 {(x-1)^{n-1}(x-2)^{n-1}\cdot (-1)} \, dx=-I_{n-1}\\ I_n \frac{4n+2}{n} =-I_{n-1}\\ (4n+2)I_n+nI_{n-1}=0[/tex]
S-a folosit integrarea prin parti?
Da, la relatiile (2) si (3).
Nu am înţeles de unde e 1/n+1 in faţa celui de-al doilea In si acel n+1 la exponent...
(x-1)^{n+1} este prin derivare (n+1)x^n. Acel 1/(n+1) are rolul de a se simplifica cu (n+1) obtinut in urma derivarii.
Ok. Mulţumesc!
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