integral from 1 to 4 of 3x+sqrt(x)

解

\int_{1}^{4} 3 x + x d x

\frac{163}{6}

十進法表記

27.16666 \dots

解答ステップ

\int_{1}^{4} 3 x + x 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}^{4} 3 x d x + \int_{1}^{4} x d x

\int_{1}^{4} 3 x d x = \frac{45}{2}

\int_{1}^{4} x d x = \frac{14}{3}

= \frac{45}{2} + \frac{14}{3}

簡素化

\frac{45}{2} + \frac{14}{3} : \frac{163}{6}

= \frac{163}{6}