integral from 1 to 3 of (2-x)^5

解

\int_{1}^{3} (2 - x)^{5} d x

0

解答ステップ

\int_{1}^{3} (2 - x)^{5} d x

拡張

(2 - x)^{5} : 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5}

= \int_{1}^{3} 32 - 80 x + 80 x^{2} - 40 x^{3} + 10 x^{4} - x^{5} 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}^{3} 32 d x - \int_{1}^{3} 80 x d x + \int_{1}^{3} 80 x^{2} d x - \int_{1}^{3} 40 x^{3} d x + \int_{1}^{3} 10 x^{4} d x - \int_{1}^{3} x^{5} d x

\int_{1}^{3} 32 d x = 64

\int_{1}^{3} 80 x d x = 320

\int_{1}^{3} 80 x^{2} d x = \frac{2080}{3}

\int_{1}^{3} 40 x^{3} d x = 800

\int_{1}^{3} 10 x^{4} d x = 484

\int_{1}^{3} x^{5} d x = \frac{364}{3}

= 64 - 320 + \frac{2080}{3} - 800 + 484 - \frac{364}{3}

簡素化

64 - 320 + \frac{2080}{3} - 800 + 484 - \frac{364}{3} : 0

= 0