integral from 0 to x^2 of e^{t^3}

解

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

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

解答ステップ

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

非初等積分を使用する:

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

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

限度を計算する:

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

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