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

Lösung

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

- 18

Schritte zur Lösung

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

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{3} x (x - 4) d x

Multipliziere aus

x (x - 4) : x^{2} - 4 x

= 2 \cdot \int_{0}^{3} x^{2} - 4 x 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

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

\int_{0}^{3} x^{2} d x = 9

\int_{0}^{3} 4 x d x = 18

= 2 (9 - 18)

Vereinfache

= - 18

\int_{4}^{\frac{1}{x}} arctan (5 t) d t

\int_{- 2}^{2} 9 - x^{2} d x

\int_{0}^{π} co x (3 x) d x

\int_{1}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{2}} d x

\int_{- 1}^{3} x^{2} - 2 x + 3 d x