integral from 0 to 1 of e^{x^3}

解答

\int_{0}^{1} e^{x^{3}} d x

- \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - (- \frac{0 \cdot Γ ( \frac{1}{3} , - 0 ^{3} )}{3 3 - 0 ^{3}})

求解步骤

\int_{0}^{1} e^{x^{3}} d x

这是非初等函数积分:

\int e^{x^{a}} d x = - \frac{x Γ ( \frac{1}{a} , - x ^{a} )}{a a - x ^{a}}

= [- \frac{x Γ ( \frac{1}{3} , - x ^{3} )}{3 3 - x ^{3}}]_{0}^{1}

带入上下限计算后得:

- \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - (- \frac{0 \cdot Γ ( \frac{1}{3} , - 0 ^{3} )}{3 3 - 0 ^{3}})

= - \frac{1 \cdot Γ ( \frac{1}{3} , - 1 ^{3} )}{3 3 - 1 ^{3}} - (- \frac{0 \cdot Γ ( \frac{1}{3} , - 0 ^{3} )}{3 3 - 0 ^{3}})

t an g e n t f (x) = \frac{- 2 x}{x ^{2} + 1}, (1, - 1)

\frac{\partial}{\partial x} ((13 - x)^{80})

y (1 + x^{2}) y^{^{'}} = x (1 + y^{2})

\frac{\partial}{\partial x} (\frac{( y - b )}{m})

x \to 0 lim (\frac{4}{x ^{2} + x} - \frac{4}{x})