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

解

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

\frac{807}{4}

十進法表記

201.75

解答ステップ

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

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

\int_{1}^{4} 3 x^{3} d x = \frac{765}{4}

\int_{1}^{4} x d x = \frac{15}{2}

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

= \frac{765}{4} + \frac{15}{2} + 3

簡素化

\frac{765}{4} + \frac{15}{2} + 3 : \frac{807}{4}

= \frac{807}{4}