integral from 1 to 3 of x^2-4x+6

解

\int_{1}^{3} x^{2} - 4 x + 6 d x

\frac{14}{3}

十進法表記

4.66666 \dots

解答ステップ

\int_{1}^{3} x^{2} - 4 x + 6 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}^{3} x^{2} d x - \int_{1}^{3} 4 x d x + \int_{1}^{3} 6 d x

\int_{1}^{3} x^{2} d x = \frac{26}{3}

\int_{1}^{3} 4 x d x = 16

\int_{1}^{3} 6 d x = 12

= \frac{26}{3} - 16 + 12

簡素化

\frac{26}{3} - 16 + 12 : \frac{14}{3}

= \frac{14}{3}