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

解

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

\frac{64}{3}

十進法表記

21.33333 \dots

解答ステップ

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

\int_{2}^{4} 2 x^{2} d x = \frac{112}{3}

\int_{2}^{4} 3 x d x = 18

\int_{2}^{4} 1 d x = 2

= \frac{112}{3} - 18 + 2

簡素化

\frac{112}{3} - 18 + 2 : \frac{64}{3}

= \frac{64}{3}