integral from-1 to 1 of x^3e^x

解答

\int_{- 1}^{1} x^{3} e^{x} d x

- 2 e + \frac{16}{e}

十进制

0.44950 \dots

求解步骤

\int_{- 1}^{1} x^{3} e^{x} d x

使用分布积分法

= [x^{3} e^{x} - \int 3 x^{2} e^{x} d x]_{- 1}^{1}

\int 3 x^{2} e^{x} d x = 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x}))

= [x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x}))]_{- 1}^{1}

带入上下限计算后得:

- 2 e + \frac{16}{e}

= - 2 e + \frac{16}{e}

\int_{0}^{2} (x^{3} + x) d x

\int_{0}^{1} 2 x (x^{2} + 3)^{3} d x

\int_{\frac{π}{2}}^{x} cos (t) d t

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

\int_{1}^{2} 3 x^{3} + 4 x^{2} - 3 x + 2 d x