integral from ln(2) to ln(3) of e^{-x}

解答

\int_{l n (2)}^{l n (3)} e^{- x} d x

\frac{1}{6}

十进制

0.16666 \dots

求解步骤

\int_{l n (2)}^{l n (3)} e^{- x} d x

使用换元积分法

= \int_{- l n (2)}^{- l n (3)} - e^{u} d u

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

= - \int_{- l n (3)}^{- l n (2)} - e^{u} d u

提出常数:

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

= - (- \int_{- l n (3)}^{- l n (2)} e^{u} d u)

使用积分常见定则:

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

= - (- [e^{u}]_{- l n (3)}^{- l n (2)})

化简

= [e^{u}]_{- l n (3)}^{- l n (2)}

带入上下限计算后得:

\frac{1}{6}

= \frac{1}{6}

\int_{1}^{e} \frac{1}{t} d t

\int_{0}^{4} e^{2 x + 1} d x

\int_{0}^{2} (2 x + 1) d x

\int_{0}^{2} 7 d x

\int_{0}^{1} x 2^{x} d x