integral of e^{(-t)/2}

解答

\int e^{\frac{- t}{2}} d t

- 2 e^{- \frac{t}{2}} + C

求解步骤

\int e^{\frac{- t}{2}} d t

化简

= \int e^{- \frac{t}{2}} d t

使用换元积分法

= \int - 2 e^{u} d u

提出常数:

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

= - 2 \cdot \int e^{u} d u

使用积分常见定则:

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

= - 2 e^{u}

u = - \frac{t}{2}

代回

= - 2 e^{- \frac{t}{2}}

解答补常数

= - 2 e^{- \frac{t}{2}} + C

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