integral from 0 to t of (t-u)

Lösung

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

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

Schritte zur Lösung

\int_{0}^{t} t - u 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} t d u - \int_{0}^{t} u d u

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

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

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

\int_{2}^{x} (\frac{1}{2} t^{2} - 1)^{8} d t

\int_{0}^{3} e^{1 - 6 t} d t

\int_{0}^{t} e^{2 u} d u

\int_{0}^{2 π} 8 sin (θ) cos (θ) d θ

\int_{1}^{7} \frac{5}{x} - \frac{5}{x ^{2}} d x