Question

utilizând metoda integrarii prin parti calculati

Answer 1

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Answer 2

$\displaystyle \mathtt{\int\limits_{-1}^2e^x(x-3)dx}\\ \\ \mathtt{f(x)=x-3\Rightarrow f'(x)=(x-3)'=x'-3'=1}\\ \\ \mathtt{g'(x)=e^x\Rightarrow g(x)=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits e^x(x-3)dx=e^x(x-3)-\int\limits e^xdx=e^x(x-3)-e^x+C=}\\ \\ \mathtt{=xe^x-3e^x-e^x+C=e^x(x-3-1)+C=e^x(x-4)+C}$
$\displaystyle \mathtt{\int\limits_{-1}^2 e^x(x-3)dx=e^x(x-4)\Bigg|_{-1}^2=e^2(2-4)-e^{-1}(-1-4)=}\\ \\ \mathtt{=e^2 \cdot (-2)- \frac{1}{e}\cdot(-5)=-2e^2+ \frac{5}{e}}$