integral from x to x+2 of (4t+1)

Lösung

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

8 x + 10

Schritte zur Lösung

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

\int_{x}^{x + 2} 4 t d t = 8 (x + 1)

\int_{x}^{x + 2} 1 d t = 2

= 8 (x + 1) + 2

Vereinfache

8 (x + 1) + 2 : 8 x + 10

= 8 x + 10