integral from-2 to 1 of x(x-1)(x+2)

解

\int_{- 2}^{1} x (x - 1) (x + 2) d x

\frac{9}{4}

十進法表記

2.25

解答ステップ

\int_{- 2}^{1} x (x - 1) (x + 2) d x

部分積分を適用する

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

\int \frac{x ^{3}}{3} - \frac{x ^{2}}{2} d x = \frac{x ^{4}}{12} - \frac{x ^{3}}{6}

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

簡素化

[\frac{1}{6} (2 x^{4} + x^{3} - 6 x^{2} - 6 (\frac{x ^{4}}{12} - \frac{x ^{3}}{6}))]_{- 2}^{1} : [\frac{1}{12} (3 x^{4} + 4 x^{3} - 12 x^{2})]_{- 2}^{1}

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

限度を計算する:

\frac{9}{4}

= \frac{9}{4}