integral from 1 to 2 of (x^2+x^{-4})

解

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

\frac{21}{8}

十進法表記

2.625

解答ステップ

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

\int_{1}^{2} x^{2} d x = \frac{7}{3}

\int_{1}^{2} x^{- 4} d x = \frac{7}{24}

= \frac{7}{3} + \frac{7}{24}

簡素化

\frac{7}{3} + \frac{7}{24} : \frac{21}{8}

= \frac{21}{8}