integral from-1 to 5 of (4x^3-3x^2+1)x

解

\int_{- 1}^{5} (4 x^{3} - 3 x^{2} + 1) x d x

\frac{10224}{5}

十進法表記

2044.8

解答ステップ

\int_{- 1}^{5} (4 x^{3} - 3 x^{2} + 1) x d x

拡張

(4 x^{3} - 3 x^{2} + 1) x : 4 x^{4} - 3 x^{3} + x

= \int_{- 1}^{5} 4 x^{4} - 3 x^{3} + 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}^{5} 4 x^{4} d x - \int_{- 1}^{5} 3 x^{3} d x + \int_{- 1}^{5} x d x

\int_{- 1}^{5} 4 x^{4} d x = \frac{12504}{5}

\int_{- 1}^{5} 3 x^{3} d x = 468

\int_{- 1}^{5} x d x = 12

= \frac{12504}{5} - 468 + 12

簡素化

\frac{12504}{5} - 468 + 12 : \frac{10224}{5}

= \frac{10224}{5}