integral from 0 to 2 of (6x-3)(4x^2+4)

Lösung

\int_{0}^{2} (6 x - 3) (4 x^{2} + 4) d x

88

Schritte zur Lösung

\int_{0}^{2} (6 x - 3) (4 x^{2} + 4) d x

Multipliziere aus

(6 x - 3) (4 x^{2} + 4) : 24 x^{3} + 24 x - 12 x^{2} - 12

= \int_{0}^{2} 24 x^{3} + 24 x - 12 x^{2} - 12 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}^{2} 24 x^{3} d x + \int_{0}^{2} 24 x d x - \int_{0}^{2} 12 x^{2} d x - \int_{0}^{2} 12 d x

\int_{0}^{2} 24 x^{3} d x = 96

\int_{0}^{2} 24 x d x = 48

\int_{0}^{2} 12 x^{2} d x = 32

\int_{0}^{2} 12 d x = 24

= 96 + 48 - 32 - 24

Vereinfache

= 88

\int_{1}^{2} \frac{5}{x ^{3}} d x

\int_{1}^{8} x e^{x^{2}} d x

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

\int_{0}^{1} e^{x} + x^{e} d x

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