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

解答

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

\frac{1}{2}

十进制

0.5

求解步骤

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

使用三角恒等式改写

= \int_{0}^{l n (2)} 2 sinh (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}^{l n (2)} sinh (x) d x

使用积分常见定则:

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

= 2 [cosh (x)]_{0}^{l n (2)}

带入上下限计算后得:

\frac{1}{4}

= 2 \cdot \frac{1}{4}

化简

= \frac{1}{2}

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