integral from 0 to t of (t-u)y(u)

解

\int_{0}^{t} (t - u) y (u) d u

y (\frac{t ^{3}}{2} - \frac{t ^{3}}{3})

解答ステップ

\int_{0}^{t} (t - u) y u d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= y \cdot \int_{0}^{t} (t - u) u d u

拡張

(t - u) u : t u - u^{2}

= y \cdot \int_{0}^{t} t u - u^{2} d u

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

= y (\int_{0}^{t} t u d u - \int_{0}^{t} u^{2} d u)

\int_{0}^{t} t u d u = \frac{t ^{3}}{2}

\int_{0}^{t} u^{2} d u = \frac{t ^{3}}{3}

= y (\frac{t ^{3}}{2} - \frac{t ^{3}}{3})