integral from-1 to 4 of (-x^2+6x+8-x^2)

解

\int_{- 1}^{4} (- x^{2} + 6 x + 8 - x^{2}) d x

\frac{125}{3}

十進法表記

41.66666 \dots

解答ステップ

\int_{- 1}^{4} - x^{2} + 6 x + 8 - x^{2} d x

簡素化

= \int_{- 1}^{4} - 2 x^{2} + 6 x + 8 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}^{4} 2 x^{2} d x + \int_{- 1}^{4} 6 x d x + \int_{- 1}^{4} 8 d x

\int_{- 1}^{4} 2 x^{2} d x = \frac{130}{3}

\int_{- 1}^{4} 6 x d x = 45

\int_{- 1}^{4} 8 d x = 40

= - \frac{130}{3} + 45 + 40

簡素化

- \frac{130}{3} + 45 + 40 : \frac{125}{3}

= \frac{125}{3}