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

解

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

\frac{48}{5}

十進法表記

9.6

解答ステップ

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

拡張

(3 x^{2} + 1)^{2} : 9 x^{4} + 6 x^{2} + 1

= \int_{- 1}^{1} 9 x^{4} + 6 x^{2} + 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_{- 1}^{1} 9 x^{4} d x + \int_{- 1}^{1} 6 x^{2} d x + \int_{- 1}^{1} 1 d x

\int_{- 1}^{1} 9 x^{4} d x = \frac{18}{5}

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

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

= \frac{18}{5} + 4 + 2

簡素化

\frac{18}{5} + 4 + 2 : \frac{48}{5}

= \frac{48}{5}