Question

Tính tích phân : \(I=\int\limits^1_0\left(x-e^{2x}\right)xdx\)

Answer 1

\(I=\int\limits^1_0\left(x+e^{2x}\right)xdx=\int\limits^1_0x^2dx+\int\limits^1_0xe^{2x}dx=I_1+I_2\)

\(I_1=\int\limits^1_0x^2dx=\frac{x^3}{3}|^1_0=\frac{1}{3}\)

Đặt \(\begin{cases}dv=e^{2x}dx\\u=x\end{cases}\) ta có \(\begin{cases}v=\frac{e^{2x}}{2}\\du=dx\end{cases}\)

\(I_2=\frac{xe^{2x}}{2}|^1_0-\int\limits^1_0\frac{e^{2x}}{2}dx=\left(\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\right)|^1_0=\frac{e^2+1}{4}\)

\(I=I_1+I_2=\frac{e^2+1}{4}+\frac{1}{3}=\frac{3e^2+7}{12}\)