integral from 1 to 6 of x^3+x^2+3

解

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

\frac{4925}{12}

十進法表記

410.41666 \dots

解答ステップ

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

\int_{1}^{6} x^{3} d x = \frac{1295}{4}

\int_{1}^{6} x^{2} d x = \frac{215}{3}

\int_{1}^{6} 3 d x = 15

= \frac{1295}{4} + \frac{215}{3} + 15

簡素化

\frac{1295}{4} + \frac{215}{3} + 15 : \frac{4925}{12}

= \frac{4925}{12}