integral from 0 to 2 of 6x^5e^{x^3y}

解

\int_{0}^{2} 6 x^{5} e^{x^{3} y} d y

6 x^{2} (e^{2 x^{3}} - 1)

解答ステップ

\int_{0}^{2} 6 x^{5} e^{x^{3} y} d y

定数を除く:

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

= 6 x^{5} \cdot \int_{0}^{2} e^{x^{3} y} d y

u置換積分法を適用する

= 6 x^{5} \cdot \int_{0}^{x^{3} \cdot 2} \frac{e ^{u}}{x ^{3}} d u

定数を除く:

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

= 6 x^{5} \frac{1}{x ^{3}} \cdot \int_{0}^{x^{3} \cdot 2} e^{u} d u

共通積分を使用する:

\int e^{u} d u = e^{u}

= 6 x^{5} \frac{1}{x ^{3}} [e^{u}]_{0}^{x^{3} \cdot 2}

簡素化

6 x^{5} \frac{1}{x ^{3}} [e^{u}]_{0}^{x^{3} \cdot 2} : 6 x^{2} [e^{u}]_{0}^{2 x^{3}}

= 6 x^{2} [e^{u}]_{0}^{2 x^{3}}

限度を計算する:

e^{2 x^{3}} - 1

= 6 x^{2} (e^{2 x^{3}} - 1)