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

解

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

\frac{78}{5}

十進法表記

15.6

解答ステップ

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

拡張

(x^{2} + 1)^{2} : x^{4} + 2 x^{2} + 1

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

\int_{- 1}^{2} x^{4} d x = \frac{33}{5}

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

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

= \frac{33}{5} + 6 + 3

簡素化

\frac{33}{5} + 6 + 3 : \frac{78}{5}

= \frac{78}{5}