integral from 0 to 2 of 6y^2-2y^4+y^3

Lösung

\int_{0}^{2} 6 y^{2} - 2 y^{4} + y^{3} d y

\frac{36}{5}

Dezimale

7.2

Schritte zur Lösung

\int_{0}^{2} 6 y^{2} - 2 y^{4} + y^{3} 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_{0}^{2} 6 y^{2} d y - \int_{0}^{2} 2 y^{4} d y + \int_{0}^{2} y^{3} d y

\int_{0}^{2} 6 y^{2} d y = 16

\int_{0}^{2} 2 y^{4} d y = \frac{64}{5}

\int_{0}^{2} y^{3} d y = 4

= 16 - \frac{64}{5} + 4

Vereinfache

16 - \frac{64}{5} + 4 : \frac{36}{5}

= \frac{36}{5}

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

\int_{0}^{π} (π - x) cos (n) x d x

\int_{0}^{\frac{5}{2}} 5 x - 2 x^{2} d x

\int_{0}^{5} e^{3 x + 1} d x

\int_{0}^{\infty} 5 e^{- 5 t} d t