integral from 1 to 5 of x^3+1/(x^3)

解

\int_{1}^{5} x^{3} + \frac{1}{x ^{3}} d x

\frac{3912}{25}

十進法表記

156.48

解答ステップ

\int_{1}^{5} x^{3} + \frac{1}{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}^{5} x^{3} d x + \int_{1}^{5} \frac{1}{x ^{3}} d x

\int_{1}^{5} x^{3} d x = 156

\int_{1}^{5} \frac{1}{x ^{3}} d x = \frac{12}{25}

= 156 + \frac{12}{25}

簡素化

156 + \frac{12}{25} : \frac{3912}{25}

= \frac{3912}{25}