integral from-1 to 1 of (4-4x^2)^2

Lösung

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

\frac{256}{15}

Dezimale

17.06666 \dots

Schritte zur Lösung

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

Multipliziere aus

(4 - 4 x^{2})^{2} : 16 - 32 x^{2} + 16 x^{4}

= \int_{- 1}^{1} 16 - 32 x^{2} + 16 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

= \int_{- 1}^{1} 16 d x - \int_{- 1}^{1} 32 x^{2} d x + \int_{- 1}^{1} 16 x^{4} d x

\int_{- 1}^{1} 16 d x = 32

\int_{- 1}^{1} 32 x^{2} d x = \frac{64}{3}

\int_{- 1}^{1} 16 x^{4} d x = \frac{32}{5}

= 32 - \frac{64}{3} + \frac{32}{5}

Vereinfache

32 - \frac{64}{3} + \frac{32}{5} : \frac{256}{15}

= \frac{256}{15}

\frac{\partial}{\partial x} (tan (\frac{u}{v}))

\frac{d}{d x} ((x + y)^{2})

\int_{0}^{1} 1 + 9 x^{2} d x

\frac{d}{d x} (d - \frac{d}{x ^{2}} - a x - b)

\frac{d}{d x} (\frac{1 + cos ( 2 x )}{2})