integral of 4x^6-2x^3+7x-4

Lösung

\int 4 x^{6} - 2 x^{3} + 7 x - 4 d x

\frac{4 x ^{7}}{7} - \frac{x ^{4}}{2} + \frac{7 x ^{2}}{2} - 4 x + C

Schritte zur Lösung

\int 4 x^{6} - 2 x^{3} + 7 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 4 x^{6} d x - \int 2 x^{3} d x + \int 7 x d x - \int 4 d x

\int 4 x^{6} d x = \frac{4 x ^{7}}{7}

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 7 x d x = \frac{7 x ^{2}}{2}

\int 4 d x = 4 x

= \frac{4 x ^{7}}{7} - \frac{x ^{4}}{2} + \frac{7 x ^{2}}{2} - 4 x

Füge eine Konstante zur Lösung hinzu

= \frac{4 x ^{7}}{7} - \frac{x ^{4}}{2} + \frac{7 x ^{2}}{2} - 4 x + C

\int_{2}^{5} \frac{1}{x ^{2}} d x

\int_{0}^{2} \frac{( 4 x + 3 )}{x ^{4}} d x

\frac{d}{d x} (a sin^{3} (\frac{x}{3}))

y^{^{'}} + \frac{y}{x} = \frac{e ^{x}}{x}

t an g e n t 5 x + 4, at x = 9