integral of (3/4 x-2/5)^2

解

\int (\frac{3}{4} x - \frac{2}{5})^{2} d x

\frac{3 x ^{3}}{16} - \frac{3 x ^{2}}{10} + \frac{4}{25} x + C

解答ステップ

\int (\frac{3}{4} x - \frac{2}{5})^{2} d x

拡張

(\frac{3}{4} x - \frac{2}{5})^{2} : \frac{9 x ^{2}}{16} - \frac{3 x}{5} + \frac{4}{25}

= \int \frac{9 x ^{2}}{16} - \frac{3 x}{5} + \frac{4}{25} 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 \frac{9 x ^{2}}{16} d x - \int \frac{3 x}{5} d x + \int \frac{4}{25} d x

\int \frac{9 x ^{2}}{16} d x = \frac{3 x ^{3}}{16}

\int \frac{3 x}{5} d x = \frac{3 x ^{2}}{10}

\int \frac{4}{25} d x = \frac{4}{25} x

= \frac{3 x ^{3}}{16} - \frac{3 x ^{2}}{10} + \frac{4}{25} x

定数を解答に追加する

= \frac{3 x ^{3}}{16} - \frac{3 x ^{2}}{10} + \frac{4}{25} x + C