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

解

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

2 (9 ln^{2} (3) - 6 ln (3) + \frac{52}{27})

十進法表記

12.39358 \dots

解答ステップ

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

部分積分を適用する

= 2 [\frac{1}{3} (e^{3 x} x^{2} - 3 \cdot \int \frac{2}{3} e^{3 x} x d x)]_{0}^{l n (3)}

\int \frac{2}{3} e^{3 x} x d x = \frac{2}{27} (3 e^{3 x} x - e^{3 x})

= 2 [\frac{1}{3} (e^{3 x} x^{2} - 3 \cdot \frac{2}{27} (3 e^{3 x} x - e^{3 x}))]_{0}^{l n (3)}

3 \cdot \frac{2}{27} (3 e^{3 x} x - e^{3 x}) = \frac{2}{9} (3 e^{3 x} x - e^{3 x})

= 2 [\frac{1}{3} (e^{3 x} x^{2} - \frac{2}{9} (3 e^{3 x} x - e^{3 x}))]_{0}^{l n (3)}

限度を計算する:

9 ln^{2} (3) - 6 ln (3) + \frac{52}{27}

= 2 (9 ln^{2} (3) - 6 ln (3) + \frac{52}{27})