integral from 0 to 2 of pi(4x^2-x^4)

Lösung

\int_{0}^{2} π (4 x^{2} - x^{4}) d x

π \frac{64}{15}

Dezimale

13.40412 \dots

Schritte zur Lösung

\int_{0}^{2} π (4 x^{2} - x^{4}) d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= π \cdot \int_{0}^{2} 4 x^{2} - x^{4} 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

= π (\int_{0}^{2} 4 x^{2} d x - \int_{0}^{2} x^{4} d x)

\int_{0}^{2} 4 x^{2} d x = \frac{32}{3}

\int_{0}^{2} x^{4} d x = \frac{32}{5}

= π (\frac{32}{3} - \frac{32}{5})

Füge

\frac{32}{3} - \frac{32}{5}

zusammen:

\frac{64}{15}

= π \frac{64}{15}

t an g e n t y = cos (x), at x = \frac{π}{4}

y^{^{''}} + y^{^{'}} - y = 0

\frac{\partial}{\partial x} (\frac{4 x - 4 y}{4 x + 4 y})

\int_{- 2}^{6} (4 x) - (x^{2} - 12) d x

(y^{2} + y \cdot x) d x - x^{2} d y = 0