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受欢迎的微积分 >

integral from 0 to ln(8) of pi(e^{-x})^2

解答

\int_{0}^{l n (8)} π (e^{- x})^{2} d x

解答

\frac{63 π}{128}

+1

十进制

1.54625 \dots

求解步骤

\int_{0}^{l n (8)} π (e^{- x})^{2} d x

提出常数:

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

= π \cdot \int_{0}^{l n (8)} (e^{- x})^{2} d x

使用换元积分法

= π \cdot \int_{1}^{\frac{1}{8}} - u d u

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

= π (- \int_{\frac{1}{8}}^{1} - u d u)

提出常数:

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

= π (- (- \int_{\frac{1}{8}}^{1} u d u))

使用幂法则

= π (- (- [\frac{u ^{2}}{2}]_{\frac{1}{8}}^{1}))

化简

= π [\frac{u ^{2}}{2}]_{\frac{1}{8}}^{1}

带入上下限计算后得:

\frac{63}{128}

= π \frac{63}{128}

化简

= \frac{63 π}{128}

作图

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