Tích phân \(\int\limits^1_0x.e^{2x}\text{d}x\) bằng
\(\dfrac{e^2+1}{2}\).\(\dfrac{e^2-1}{2}\).\(\dfrac{e^2+1}{4}\).\(\dfrac{e^2-1}{4}\).Hướng dẫn giải:\(\int\limits^1_0x.e^{2x}\text{d}x=\int\limits^1_0x\left(\dfrac{e^{2x}}{2}\right)'\text{d}x=\dfrac{xe^{2x}}{2}|_0^1-\int\limits^1_0\dfrac{e^{2x}}{2}\left(x\right)'\text{d}x=\dfrac{e^2}{2}-\int\limits^1_0\dfrac{e^{2x}}{2}\text{d}x\)
\(=\dfrac{e^2}{2}-\dfrac{e^{2x}}{4}|_0^1=\dfrac{e^2}{2}-\left(\dfrac{e^2}{4}-\dfrac{1}{4}\right)=\dfrac{e^2+1}{4}\)