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

解

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

\frac{181}{9}

十進法表記

20.11111 \dots

解答ステップ

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

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} \frac{1}{4 x ^{3}} d x = \frac{1}{9}

= 20 + \frac{1}{9}

簡素化

20 + \frac{1}{9} : \frac{181}{9}

= \frac{181}{9}