Question

Integrala de la 0la 1(x la a 2a+x)e la x dx

Answer 1

$\displaystyle \mathtt{\int\limits_0^1(x^2+x)e^xdx}\\ \\ \mathtt{-----}\\ \\ \mathtt{\int\limits(x^2+x)e^xdx}\\\\\mathtt{f(x)=x^2+x~~~~~~~~~~~~~~~f'(x)=(x^2+x)'=(x^2)'+x'=2x+1}\\ \\ \mathtt{g'(x)=e^x~~~~~~~~~~~~~~~~~~g(x)=\int\limits e^xdx=e^x}\\ \\ \mathtt{\int\limits(x^2+x)e^xdx=(x^2+x)e^x-\int\limits(2x+1)e^xdx}\\ \\ \mathtt{\int\limits(2x+1)e^xdx}\\ \\ \mathtt{f(x)=2x+1~~~~~~~~~~~~~~~f'(x)=(2x+1)'=(2x)'+1'=2x'+0=2}\\ \\ \mathtt{g'(x)=e^x~~~~~~~~~~~~~~~~~~~g(x)=\int\limits e^xdx=e^x}$

$\displaystyle \mathtt{\int\limits(2x+1)e^xdx=(2x+1)e^x-\int\limits2e^xdx=(2x+1)e^x-2\int\limits e^xdx=} \\ \\ \mathtt{=(2x+1)e^x-2e^x+C=2xe^x+e^x-2e^x+C=2xe^x-e^x+C}\\ \\ \mathtt{\int\limits(x^2+x)e^xdx=(x^2+x)e^x-\int\limits(2x+1)e^xdx=}\\ \\ \mathtt{=(x^2+x)e^x-(2xe^x-e^x)+C=x^2e^x+xe^x-2xe^x+e^x+C=}\\ \\ \mathtt{=x^2e^x-xe^x+e^x+C=e^x(x^2-x+1)+C}$

$\displaystyle \mathtt{\int\limits_0^1(x^2+x)e^xdx=e^x(x^2-x+1)\Bigg|_0^1=e^1(1^2-1+1)-e^0(0^2-0+1)=}\\ \\ \mathtt{=e \cdot 1-1\cdot1=\boxed{\mathtt{e-1}}}$