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

解

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

\frac{16}{15}

十進法表記

1.06666 \dots

解答ステップ

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

拡張

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

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

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

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

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

= \frac{2}{5} - \frac{4}{3} + 2

簡素化

\frac{2}{5} - \frac{4}{3} + 2 : \frac{16}{15}

= \frac{16}{15}