integral from 0 to 1 of (1+sqrt(x))^4

解

\int_{0}^{1} (1 + x)^{4} d x

\frac{43}{5}

十進法表記

8.6

解答ステップ

\int_{0}^{1} (1 + x)^{4} d x

拡張

(1 + x)^{4} : 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2}

= \int_{0}^{1} 1 + 4 x + 6 x + 4 x^{\frac{3}{2}} + x^{2} 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_{0}^{1} 1 d x + \int_{0}^{1} 4 x d x + \int_{0}^{1} 6 x d x + \int_{0}^{1} 4 x^{\frac{3}{2}} d x + \int_{0}^{1} x^{2} d x

\int_{0}^{1} 1 d x = 1

\int_{0}^{1} 4 x d x = \frac{8}{3}

\int_{0}^{1} 6 x d x = 3

\int_{0}^{1} 4 x^{\frac{3}{2}} d x = \frac{8}{5}

\int_{0}^{1} x^{2} d x = \frac{1}{3}

= 1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3}

簡素化

1 + \frac{8}{3} + 3 + \frac{8}{5} + \frac{1}{3} : \frac{43}{5}

= \frac{43}{5}