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

解

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

\frac{15}{4}

十進法表記

3.75

解答ステップ

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

拡張

\frac{3 x ^{3} + 2}{x ^{3}} : 3 + \frac{2}{x ^{3}}

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

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

\int_{1}^{2} \frac{2}{x ^{3}} d x = \frac{3}{4}

= 3 + \frac{3}{4}

簡素化

3 + \frac{3}{4} : \frac{15}{4}

= \frac{15}{4}