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

解

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

\frac{8}{3}

十進法表記

2.66666 \dots

解答ステップ

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

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

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

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

\int_{- 1}^{1} x^{3} d x = 0

= 2 + 0 + \frac{2}{3} + 0

簡素化

2 + 0 + \frac{2}{3} + 0 : \frac{8}{3}

= \frac{8}{3}