integral from-1 to 1 of (1+6xy^2)

Lösung

\int_{- 1}^{1} (1 + 6 x y^{2}) d y

2 + 4 x

Schritte zur Lösung

\int_{- 1}^{1} 1 + 6 x y^{2} d y

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}^{1} 1 d y + \int_{- 1}^{1} 6 x y^{2} d y

\int_{- 1}^{1} 1 d y = 2

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

= 2 + 4 x

\int_{0}^{4} - 6 x^{2} - 12 x + 144 d x

\int_{0}^{b} 7 e^{x} d x

\int_{0}^{x} cos (9 t) d t

\int_{0}^{6} x^{2} - 3 x d x

\int_{0}^{3} t e^{- 0.6 t} d t