integral from 0 to 1 of x(e^x)

解答

\int_{0}^{1} x (e^{x}) d x

1

求解步骤

\int_{0}^{1} x e^{x} d x

使用分布积分法

= [e^{x} x - \int e^{x} d x]_{0}^{1}

\int e^{x} d x = e^{x}

= [e^{x} x - e^{x}]_{0}^{1}

带入上下限计算后得:

1

= 1

\int_{1}^{3} y^{2} d y

\int_{1}^{2} (2 x + \frac{2}{x ^{2}}) d x

\int_{0}^{1} 7 + 4 t^{2} d t

\int_{0}^{2} y (2 y^{2} - \frac{y ^{2}}{2}) d y

\int_{4}^{5} x^{2} + 2 x + 2 d x