integral from 0 to 5 of xe^{-4x}

解答

\int_{0}^{5} x e^{- 4 x} d x

- \frac{- e ^{20} + 21}{16 e ^{20}}

十进制

0.06249 \dots

求解步骤

\int_{0}^{5} x e^{- 4 x} d x

使用换元积分法

= \int_{0}^{- 20} \frac{e ^{u} u}{16} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 20}^{0} \frac{e ^{u} u}{16} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{16} \cdot \int_{- 20}^{0} e^{u} u d u

使用分布积分法

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 20}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 20}^{0}

带入上下限计算后得:

- 1 + \frac{21}{e ^{20}}

= - \frac{1}{16} (- 1 + \frac{21}{e ^{20}})

化简

= - \frac{- e ^{20} + 21}{16 e ^{20}}

\int \frac{2}{9 - 4 x ^{2}} d x

\int_{1}^{3} \frac{1}{( x - 3 ) ^{2}} d x

\int 7 v (v^{2} + 4)^{2} d v

\int 2 + 2 sin (x) d x

\int 2 + \frac{1}{x} d x