integral from 1 to 5 of (2x+1)

Lösung

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

28

Schritte zur Lösung

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

\int_{1}^{5} 2 x d x = 24

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

= 24 + 4

Vereinfache

= 28

\int_{- 1}^{0} - 20 x e^{- x} d x

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

\int_{0}^{1} c x^{2} d x

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

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