integral from 1 to 3 of (2-x)^2ln(4x)

解

\int_{1}^{3} (2 - x)^{2} ln (4 x) d x

3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)

十進法表記

1.33114 \dots

解答ステップ

\int_{1}^{3} (2 - x)^{2} ln (4 x) d x

部分積分を適用する

= [ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - \int \frac{1}{3} (12 - 6 x + x^{2}) d x]_{1}^{3}

\int \frac{1}{3} (12 - 6 x + x^{2}) d x = 4 x - x^{2} + \frac{x ^{3}}{9}

= [ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - (4 x - x^{2} + \frac{x ^{3}}{9})]_{1}^{3}

簡素化

[ln (4 x) (4 x - 2 x^{2} + \frac{x ^{3}}{3}) - (4 x - x^{2} + \frac{x ^{3}}{9})]_{1}^{3} : [4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9}]_{1}^{3}

= [4 x ln (4 x) - 2 x^{2} ln (4 x) + \frac{1}{3} x^{3} ln (4 x) - 4 x + x^{2} - \frac{x ^{3}}{9}]_{1}^{3}

限度を計算する:

3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)

= 3 ln (12) - \frac{26}{9} - \frac{14}{3} ln (2)