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

Lösung

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

6

Schritte zur Lösung

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

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

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

= 3 + 3

Vereinfache

= 6

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

\int_{1}^{x} \frac{e ^{t}}{t} d t

\int_{0}^{1} (1 - x)^{5} d x

\int_{0}^{1} (1 - y)^{2} d y

\int_{0}^{6 - x} 6 - x - z d z