integral from-1 to 1 of (\sqrt[3]{t}-6)

Lösung

\int_{- 1}^{1} (3 t - 6) d t

- 12

Schritte zur Lösung

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

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

\int_{- 1}^{1} 6 d t = 12

= 0 - 12

Vereinfache

= - 12

\int_{5}^{x^{2}} \frac{1}{t ^{4}} d t

\int_{0}^{x} 3 t^{2} d t

\int_{0}^{\frac{7}{3}} 3 t - 7 d t

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

\int_{3}^{10} 5 d x