Question

Calculati $\int\limits^1_0 {x e^{-x} , dx$ . Va multumesc anticipat!

Answer 1

faci prin parti 

[tex] \int\limits^0_1 {x* e^{-x} } \, dx =} \, -\int\limits^0_1 {x* (e^{-x})^' } \, dx =} -\int\limits^0_1 {(x )'* e^{-x} } \, dx } - (e^ {-x} *x)| _{0} ^{1} = \\ =-\int\limits^0_1 { e^{-x} } \, dx } -e^{-1}=-e^ {-x} | _{0} ^{1}-e^{-1}=-2*e^{-1}+1[/tex]

Answer 2

$\displaystyle \mathtt{\int\limits_0^1xe^{-x}dx}\\ \\ \mathtt{\int\limits f(x)g'(x)dx=f(x)g(x)-\int\limits f'(x)g(x)dx}\\ \\ \mathtt{f(x)=x\Rightarrow f'(x)=x'=1}\\ \\ \mathtt{g'(x)=e^{-x}\Rightarrow g(x)=\int\limits e^{-x}dx=-e^{-x}}\\ \\ \mathtt{\int\limits xe^{-x}dx=-xe^{-x}+\int\limits e^{-x}dx=-xe^{-x}-e^{-x}+C=-e^{-x}(x+1)+C}\\ \\ \mathtt{\int\limits_0^1xe^{-x}dx=-e^{-x}(x+1)\Bigg|_0^1=-e^{-1}(1+1)+e^0(0+1)=}\\ \\ \mathtt{=-2e^{-1}+1 \cdot 1=-2 \cdot \frac{1}{e}+1= \frac{-2}{e}+1= \frac{-2+e}{e} }$