integral from 1 to t of-5t+9

Lösung

\int_{1}^{t} - 5 t + 9 d t

- \frac{5 ( t ^{2} - 1 )}{2} + 9 t - 9

Schritte zur Lösung

\int_{1}^{t} - 5 t + 9 d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{1}^{t} 5 t d t + \int_{1}^{t} 9 d t

\int_{1}^{t} 5 t d t = \frac{5 ( t ^{2} - 1 )}{2}

\int_{1}^{t} 9 d t = 9 t - 9

= - \frac{5 ( t ^{2} - 1 )}{2} + 9 t - 9

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