integral from-1 to 1 of (\sqrt[3]{t}-9)

Lösung

\int_{- 1}^{1} (3 t - 9) d t

- 18

Schritte zur Lösung

\int_{- 1}^{1} 3 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}^{1} 3 t d t - \int_{- 1}^{1} 9 d t

\int_{- 1}^{1} 3 t d t = 0

\int_{- 1}^{1} 9 d t = 18

= 0 - 18

Vereinfache

= - 18

\int_{- 2}^{1} (2 x^{2} + 2) d x

\int_{0}^{1.25} x^{3} + 2 x^{2} - 3 d x

\int_{0}^{y^{3}} e^{y^{4}} d y

\int_{0}^{\frac{π}{2}} sin^{2} (2 t) d t

\int_{0}^{11} x d x