integral of (x^2-3x-4)

Lösung

\int (x^{2} - 3 x - 4) d x

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

Schritte zur Lösung

\int x^{2} - 3 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 x^{2} d x - \int 3 x d x - \int 4 d x

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

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

\int 4 d x = 4 x

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

Füge eine Konstante zur Lösung hinzu

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

t an g e n t f (x) = x + 4, (12, 4)

4 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

d er i v a t i v e f (x) = \frac{x ^{3}}{6 x - 3}

x \to 3 + lim (\frac{2 + x}{3 - x})

\frac{d y}{d t} = \frac{t ^{2} + t y + y ^{2}}{t ^{2}}