Question

Tính tích phân :

\(I=\int^{\ln3}_1\left(x^2-2x\right)e^xdx\)

 

 

Answer 1

\(I=\int\limits^{\ln3}_1\left(x^2-2x\right)de^x=\left(x^2-2x\right)e^x|^{\ln3}_1-\int\limits_1^{\ln3}e^xd\left(x^2-2x\right)=3\left(\ln^23-2\ln3\right)+e-2\int\limits^{\ln3}_1\left(x-1\right)e^xdx\)

\(\int\limits^{\ln3}_1\left(x-1\right)e^xdx=k\)

Lại có :

\(k=\int\limits^{\ln3}_1\left(x-1\right)de^x=\left(x-1\right)e^x|^{\ln3}_0-\int\limits^{\ln3}_0e^xd\left(x-1\right)=3\left(\ln3-1\right)-e^x|^{\ln3}_0=3\ln3-6+e\)

Do đó :

\(I=3\left(\ln^23-2\ln3\right)+e-2\left(3\ln3-6+e\right)=3\ln^23-12\ln3+12-e\)