integral from 0 to t of (y(u)-1)

Lösung

\int_{0}^{t} (y (u) - 1) d u

- t + \frac{y t ^{2}}{2}

Schritte zur Lösung

\int_{0}^{t} y u - 1 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

= - \int_{0}^{t} 1 d u + \int_{0}^{t} y u d u

\int_{0}^{t} 1 d u = t

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

= - t + \frac{y t ^{2}}{2}

\int_{0}^{2} 12 x^{3} x^{4} + 9 d x

\int_{0}^{2} 2 cos (\frac{nπ x}{2}) d x

\int_{1}^{4} 8 x d x

\int_{0}^{2} 3 x (x - 7) d x

\int_{0}^{1} 2 π x e^{x} d x