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

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Multipliziere aus

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

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

Wende die Summenregel an:

\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})

\int_{0}^{e} \frac{x ^{3}}{1 + x ^{2}} d x

\int_{0}^{4} 48 - 3 x^{2} d x

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