integral from 0 to 4 of xe^x

解答

\int_{0}^{4} x e^{x} d x

3 e^{4} + 1

十进制

164.79445 \dots

求解步骤

\int_{0}^{4} x e^{x} d x

使用分布积分法

= [e^{x} x - \int e^{x} d x]_{0}^{4}

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

= [e^{x} x - e^{x}]_{0}^{4}

带入上下限计算后得:

3 e^{4} + 1

= 3 e^{4} + 1

(ln (x^{2} + 1))^{^{'}}

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (- 1) = 0, y^{^{'}} (- 1) = 1

\int \frac{3 x + 5}{x ( 2 x - 1 )} d x

d er i v a t i v e t^{3} e^{t} t

n \to \infty lim (c^{n})