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

解

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

\frac{7}{6}

十進法表記

1.16666 \dots

解答ステップ

\int_{1}^{2} 1 - 3 x + 2 x^{2} 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_{1}^{2} 1 d x - \int_{1}^{2} 3 x d x + \int_{1}^{2} 2 x^{2} d x

\int_{1}^{2} 1 d x = 1

\int_{1}^{2} 3 x d x = \frac{9}{2}

\int_{1}^{2} 2 x^{2} d x = \frac{14}{3}

= 1 - \frac{9}{2} + \frac{14}{3}

簡素化

1 - \frac{9}{2} + \frac{14}{3} : \frac{7}{6}

= \frac{7}{6}