integral of (x^4-1/2 x^3+1/4 x-2)

解

\int (x^{4} - \frac{1}{2} x^{3} + \frac{1}{4} x - 2) d x

\frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x + C

解答ステップ

\int x^{4} - \frac{1}{2} x^{3} + \frac{1}{4} 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 x^{4} d x - \int \frac{1}{2} x^{3} d x + \int \frac{1}{4} x d x - \int 2 d x

\int x^{4} d x = \frac{x ^{5}}{5}

\int \frac{1}{2} x^{3} d x = \frac{x ^{4}}{8}

\int \frac{1}{4} x d x = \frac{x ^{2}}{8}

\int 2 d x = 2 x

= \frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x

定数を解答に追加する

= \frac{x ^{5}}{5} - \frac{x ^{4}}{8} + \frac{x ^{2}}{8} - 2 x + C