integral from 0 to t of 20te^{-10t}

解答

\int_{0}^{t} 20 t e^{- 10 t} d t

\frac{1}{5} (- 10 e^{- 10 t} t - e^{- 10 t} + 1)

求解步骤

\int_{0}^{t} 20 t e^{- 10 t} d t

提出常数:

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

= 20 \cdot \int_{0}^{t} t e^{- 10 t} d t

使用换元积分法

= 20 \cdot \int_{0}^{- 10 t} \frac{e ^{u} u}{100} d u

提出常数:

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

= 20 \cdot \frac{1}{100} \cdot \int_{0}^{- 10 t} e^{u} u d u

使用分布积分法

= 20 \cdot \frac{1}{100} [e^{u} u - \int e^{u} d u]_{0}^{- 10 t}

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

= 20 \cdot \frac{1}{100} [e^{u} u - e^{u}]_{0}^{- 10 t}

化简

20 \cdot \frac{1}{100} [e^{u} u - e^{u}]_{0}^{- 10 t} : \frac{1}{5} [e^{u} u - e^{u}]_{0}^{- 10 t}

= \frac{1}{5} [e^{u} u - e^{u}]_{0}^{- 10 t}

带入上下限计算后得:

- 10 e^{- 10 t} t - e^{- 10 t} + 1

= \frac{1}{5} (- 10 e^{- 10 t} t - e^{- 10 t} + 1)

\int_{\frac{π}{4}}^{\frac{π}{2}} csc (x) d x

\int_{0}^{4} \frac{x}{1 + x} d x

\int_{1}^{7} \frac{9}{1 + x ^{2}} d x

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

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