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

解

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

e^{2} 4 e (π erfi (\frac{1}{2}) - π erfi (- \frac{3}{2}))

解答ステップ

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

定数を除く:

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

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

平方完成する

x^{2} - 3 x : (x - \frac{3}{2})^{2} - \frac{9}{4}

= 2 \cdot \int_{0}^{2} e^{(x - \frac{3}{2})^{2} - \frac{9}{4}} d x

u置換積分法を適用する

= 2 \cdot \int_{- \frac{3}{2}}^{\frac{1}{2}} e^{u^{2} - \frac{9}{4}} d u

簡素化

e^{u^{2} - \frac{9}{4}} : e^{u^{2}} 4 e^{9}

= 2 \cdot \int_{- \frac{3}{2}}^{\frac{1}{2}} e^{u^{2}} 4 e^{9} d u

定数を除く:

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

= 2 4 e^{9} \cdot \int_{- \frac{3}{2}}^{\frac{1}{2}} e^{u^{2}} d u

4 e^{9} = e^{2} 4 e

= 2 e^{2} 4 e \cdot \int_{- \frac{3}{2}}^{\frac{1}{2}} e^{u^{2}} d u

非初等積分を使用する:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= 2 e^{2} 4 e [\frac{π}{2} erfi (u)]_{- \frac{3}{2}}^{\frac{1}{2}}

限度を計算する:

\frac{π erfi ( \frac{1}{2} ) - π erfi ( - \frac{3}{2} )}{2}

= 2 e^{2} 4 e \frac{π erfi ( \frac{1}{2} ) - π erfi ( - \frac{3}{2} )}{2}

簡素化

= e^{2} 4 e (π erfi (\frac{1}{2}) - π erfi (- \frac{3}{2}))