integral from 0 to 10 of t+4

Lösung

\int_{0}^{10} t + 4 d t

90

Schritte zur Lösung

\int_{0}^{10} t + 4 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_{0}^{10} t d t + \int_{0}^{10} 4 d t

\int_{0}^{10} t d t = 50

\int_{0}^{10} 4 d t = 40

= 50 + 40

Vereinfache

= 90

\int_{0}^{1} \frac{( 1 - x ) ^{2}}{2} d x

\int_{1}^{\infty} \frac{1}{2 x + 2} d x

\int_{- 2}^{2} 2 x^{2} - (x^{4} - 2 x^{2}) d x

\int_{0}^{x} 3 y d y

\int_{- r}^{r} x r^{2} - x^{2} d x