Question

Ma poate ajuta cineva cu acest exercițiu

Answer 1

Explicație pas cu pas:

$I_{2} = \int_{0}^{1} \ 26x^{2}(1-x)^{23} \ dx = -26 \int_{0}^{1} \ (x-1)^{23}x^{2} \ dx$

notăm

$x-1 = u \to dx = du \implies x^{2} = (u+1)^{2}$

$\int \ u^{23}(u+1)^{2} \ du = \int \ u^{23}(u^{2}+2u+1) \ du =$

$= \int \ u^{23} \cdot u^{2} \ du + \int \ u^{23} \cdot 2u \ du + \int \ u^{23} \ du$

$= \int \ u^{25} \ du + 2\int \ u^{24} \ du + \int \ u^{23} \ du$

$= \dfrac{u^{26}}{26} + \dfrac{2u^{25}}{25} + \dfrac{u^{24}}{24}$

$I_{2} = -26 \bigg(\dfrac{(x-1)^{26}}{26} + \dfrac{2(x-1)^{25}}{25} + \dfrac{(x-1)^{24}}{24}\bigg) + C$

$= -(x-1)^{26} - \dfrac{52(x-1)^{25}}{25} - \dfrac{13(x-1)^{24}}{12} + C$

$\int_{0}^{1} \ 26x^{2}(1-x)^{23} \ dx = - (x-1)^{26}\Big|_{0}^{1} - \dfrac{52(x-1)^{25}}{25}\Big|_{0}^{1} - \dfrac{13(x-1)^{24}}{12}\Big|_{0}^{1}$

$= -(1-1)^{26} + (0-1)^{26} - \dfrac{52(1-1)^{25}}{25} + \dfrac{52(0-1)^{25}}{25} - \dfrac{13(1-1)^{24}}{12} + \dfrac{13(0-1)^{24}}{12}$

$= (-1)^{26} + \dfrac{52(-1)^{25}}{25} + \dfrac{13(-1)^{24}}{12}$

$= 1 - \dfrac{52}{25} + \dfrac{13}{12} = \dfrac{1}{300}$