integral from 0 to 4 of 8x-32

Lösung

\int_{0}^{4} 8 x - 32 d x

- 64

Schritte zur Lösung

\int_{0}^{4} 8 x - 32 d x

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}^{4} 8 x d x - \int_{0}^{4} 32 d x

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 32 d x = 128

= 64 - 128

Vereinfache

= - 64

\int_{- π}^{π} - sin (x) d x

\int_{0}^{8} (- x^{2} + 8 x) d x

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

\int_{0}^{2 π} (\frac{π - x}{2}) d x

\int_{0}^{1} x x^{2} + 5 d x