integral from 0.1 to 0.25 of (x^3+1)/6

解

\int_{0.1}^{0.25} \frac{x ^{3} + 1}{6} d x

0.02515 \dots

解答ステップ

\int_{0.1}^{0.25} \frac{x ^{3} + 1}{6} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int_{0.1}^{0.25} x^{3} + 1 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

= \frac{1}{6} (\int_{0.1}^{0.25} x^{3} d x + \int_{0.1}^{0.25} 1 d x)

\int_{0.1}^{0.25} x^{3} d x = 0.0009515625

\int_{0.1}^{0.25} 1 d x = 0.15

= \frac{1}{6} (0.0009515625 + 0.15)

簡素化

\frac{1}{6} (0.0009515625 + 0.15) : 0.02515 \dots

= 0.02515 \dots