integral from-3 to-1 of (2t+5)

Lösung

\int_{- 3}^{- 1} (2 t + 5) d t

2

Schritte zur Lösung

\int_{- 3}^{- 1} 2 t + 5 d t

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- 3}^{- 1} 2 t d t + \int_{- 3}^{- 1} 5 d t

\int_{- 3}^{- 1} 2 t d t = - 8

\int_{- 3}^{- 1} 5 d t = 10

= - 8 + 10

Vereinfache

= 2

\int_{- \infty}^{0} x e^{\frac{x}{2}} d x

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

\int_{0}^{4} x x + 1 d x

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

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