integral from 0 to 1/2 of 2x+3-xe^{3x}

解

\int_{0}^{\frac{1}{2}} 2 x + 3 - x e^{3 x} d x

\frac{7}{4} - \frac{e ^{\frac{3}{2}} + 2}{18}

十進法表記

1.38990 \dots

解答ステップ

\int_{0}^{\frac{1}{2}} 2 x + 3 - x e^{3 x} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{\frac{1}{2}} 2 x d x + \int_{0}^{\frac{1}{2}} 3 d x - \int_{0}^{\frac{1}{2}} x e^{3 x} d x

\int_{0}^{\frac{1}{2}} 2 x d x = \frac{1}{4}

\int_{0}^{\frac{1}{2}} 3 d x = \frac{3}{2}

\int_{0}^{\frac{1}{2}} x e^{3 x} d x = \frac{e ^{\frac{3}{2}} + 2}{18}

= \frac{1}{4} + \frac{3}{2} - \frac{e ^{\frac{3}{2}} + 2}{18}

分数を組み合わせる

\frac{1}{4} + \frac{3}{2} : \frac{7}{4}

= \frac{7}{4} - \frac{e ^{\frac{3}{2}} + 2}{18}