integral from 0 to 4 of e^x+e^{-x}

解答

\int_{0}^{4} e^{x} + e^{- x} d x

\frac{e ^{8} - 1}{e ^{4}}

十进制

54.57983 \dots

求解步骤

\int_{0}^{4} e^{x} + e^{- x} d x

使用三角恒等式改写

= \int_{0}^{4} 2 cosh (x) d x

提出常数:

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

= 2 \cdot \int_{0}^{4} cosh (x) d x

使用积分常见定则:

\int cosh (x) d x = sinh (x)

= 2 [sinh (x)]_{0}^{4}

带入上下限计算后得:

\frac{e ^{8} - 1}{2 e ^{4}}

= 2 \cdot \frac{e ^{8} - 1}{2 e ^{4}}

化简

= \frac{e ^{8} - 1}{e ^{4}}

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