integral of (4-x)^2

Lösung

\int (4 - x)^{2} d x

16 x - 4 x^{2} + \frac{x ^{3}}{3} + C

Schritte zur Lösung

\int (4 - x)^{2} d x

Multipliziere aus

(4 - x)^{2} : 16 - 8 x + x^{2}

= \int 16 - 8 x + x^{2} 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 16 d x - \int 8 x d x + \int x^{2} d x

\int 16 d x = 16 x

\int 8 x d x = 4 x^{2}

\int x^{2} d x = \frac{x ^{3}}{3}

= 16 x - 4 x^{2} + \frac{x ^{3}}{3}

Füge eine Konstante zur Lösung hinzu

= 16 x - 4 x^{2} + \frac{x ^{3}}{3} + C

d er i v a t i v e e^{\frac{- x ^{2}}{2}}

\int_{0}^{1} - \frac{6 y}{e ^{2 y}} d y

x \to 3 lim (e^{\frac{7}{( x - 3 )}})

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