integral from 0 to 1 of (3x^3-x^2+2x-4)

解

\int_{0}^{1} (3 x^{3} - x^{2} + 2 x - 4) d x

- \frac{31}{12}

十進法表記

- 2.58333 \dots

解答ステップ

\int_{0}^{1} 3 x^{3} - x^{2} + 2 x - 4 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_{0}^{1} 3 x^{3} d x - \int_{0}^{1} x^{2} d x + \int_{0}^{1} 2 x d x - \int_{0}^{1} 4 d x

\int_{0}^{1} 3 x^{3} d x = \frac{3}{4}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

\int_{0}^{1} 2 x d x = 1

\int_{0}^{1} 4 d x = 4

= \frac{3}{4} - \frac{1}{3} + 1 - 4

簡素化

\frac{3}{4} - \frac{1}{3} + 1 - 4 : - \frac{31}{12}

= - \frac{31}{12}