integral from 1 to 2 of 10x^3+6/(x^2)

解

\int_{1}^{2} 10 x^{3} + \frac{6}{x ^{2}} d x

\frac{81}{2}

十進法表記

40.5

解答ステップ

\int_{1}^{2} 10 x^{3} + \frac{6}{x ^{2}} 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} 10 x^{3} d x + \int_{1}^{2} \frac{6}{x ^{2}} d x

\int_{1}^{2} 10 x^{3} d x = \frac{75}{2}

\int_{1}^{2} \frac{6}{x ^{2}} d x = 3

= \frac{75}{2} + 3

簡素化

\frac{75}{2} + 3 : \frac{81}{2}

= \frac{81}{2}