integral from 0 to 3x of (2e^{3+x^3})

解答

\int_{0}^{3 x} (2 e^{3 + x^{3}}) d x

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

求解步骤

\int_{0}^{3 x} 2 e^{3 + x^{3}} d x

提出常数:

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

= 2 \cdot \int_{0}^{3 x} e^{3 + x^{3}} d x

e^{3 + x^{3}} = e^{3} e^{x^{3}}

= 2 \cdot \int_{0}^{3 x} e^{3} e^{x^{3}} d x

提出常数:

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

= 2 e^{3} \cdot \int_{0}^{3 x} 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}}

= 2 e^{3} [- \frac{x Γ ( \frac{1}{3} , - x ^{3} )}{3 3 - x ^{3}}]_{0}^{3 x}

带入上下限计算后得:

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

= 2 e^{3} - \frac{3 x Γ ( \frac{1}{3} , - ( 3 x ) ^{3} )}{3 3 - ( 3 x ) ^{3}} - (- \frac{0 \cdot Γ ( \frac{1}{3} , - 0 ^{3} )}{3 3 - 0 ^{3}})

\int_{0}^{c} 2 e^{x} d x

\int_{1}^{e^{3}} \frac{2}{x} d x

\int_{0}^{x^{3}} e^{x^{2}} d x

\int_{- 2}^{4} (x) d x

\int_{0}^{\infty} x e^{- y x} d x