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

Lösung

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

- 21

Schritte zur Lösung

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Multipliziere aus

(2 x + 1)^{2} : 4 x^{2} + 4 x + 1

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

\int_{- 1}^{2} 4 x^{2} d x = 12

\int_{- 1}^{2} 4 x d x = 6

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

= - (12 + 6 + 3)

Vereinfache

= - 21

\int_{0}^{\frac{π}{2}} (cos (2 x)) d x

\int_{1}^{2} t 9 + 16 t^{2} d t

\int_{- 2}^{x} (4 t^{2} - 3 t) d t

\int_{0}^{1} \frac{1}{x ^{3} + x ^{2} + 2 x + 2} d x

\int_{0}^{\frac{π}{2}} 2 π x cos (x) d x