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

解

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

\frac{17}{6}

十進法表記

2.83333 \dots

解答ステップ

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

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

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

= \frac{7}{3} + \frac{3}{2} - 1

簡素化

\frac{7}{3} + \frac{3}{2} - 1 : \frac{17}{6}

= \frac{17}{6}