integral from 0 to ln(2) of xe^{-4x}

解答

\int_{0}^{l n (2)} x e^{- 4 x} d x

- \frac{1}{16} (- \frac{15}{16} + \frac{1}{4} ln (2))

十进制

0.04776 \dots

求解步骤

\int_{0}^{l n (2)} x e^{- 4 x} d x

使用换元积分法

= \int_{0}^{- 4 l n (2)} \frac{e ^{u} u}{16} d u

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

= - \int_{- 4 l n (2)}^{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_{- 4 l n (2)}^{0} e^{u} u d u

使用分布积分法

= - \frac{1}{16} [e^{u} u - \int e^{u} d u]_{- 4 l n (2)}^{0}

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

= - \frac{1}{16} [e^{u} u - e^{u}]_{- 4 l n (2)}^{0}

带入上下限计算后得:

- \frac{15}{16} + \frac{1}{4} ln (2)

= - \frac{1}{16} (- \frac{15}{16} + \frac{1}{4} ln (2))

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