integral from 1 to 4 of (x^2+2x-5)

Lösung

\int_{1}^{4} (x^{2} + 2 x - 5) d x

21

Schritte zur Lösung

\int_{1}^{4} x^{2} + 2 x - 5 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_{1}^{4} x^{2} d x + \int_{1}^{4} 2 x d x - \int_{1}^{4} 5 d x

\int_{1}^{4} x^{2} d x = 21

\int_{1}^{4} 2 x d x = 15

\int_{1}^{4} 5 d x = 15

= 21 + 15 - 15

Vereinfache

= 21

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

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

\int_{a}^{b} 0 d x

\int_{0}^{l} sin (\frac{nπ x}{l}) d x

\int_{0}^{5 π} ∣ sin (x) ∣ d x