integral from 1 to 4 of (1/(x^3)+x)

解

\int_{1}^{4} (\frac{1}{x ^{3}} + x) d x

\frac{255}{32}

十進法表記

7.96875

解答ステップ

\int_{1}^{4} \frac{1}{x ^{3}} + x 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} \frac{1}{x ^{3}} d x + \int_{1}^{4} x d x

\int_{1}^{4} \frac{1}{x ^{3}} d x = \frac{15}{32}

\int_{1}^{4} x d x = \frac{15}{2}

= \frac{15}{32} + \frac{15}{2}

簡素化

\frac{15}{32} + \frac{15}{2} : \frac{255}{32}

= \frac{255}{32}