integral of (-1/4 x^3+3/4 x^2+4)

Lösung

\int (- \frac{1}{4} x^{3} + \frac{3}{4} x^{2} + 4) d x

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

Schritte zur Lösung

\int - \frac{1}{4} x^{3} + \frac{3}{4} x^{2} + 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 \frac{1}{4} x^{3} d x + \int \frac{3}{4} x^{2} d x + \int 4 d x

\int \frac{1}{4} x^{3} d x = \frac{x ^{4}}{16}

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

\int 4 d x = 4 x

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

Füge eine Konstante zur Lösung hinzu

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

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