integral of 3/4 t+6

Lösung

\int \frac{3}{4} t + 6 d t

\frac{3 t ^{2}}{8} + 6 t + C

Schritte zur Lösung

\int \frac{3}{4} t + 6 d t

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{3}{4} t d t + \int 6 d t

\int \frac{3}{4} t d t = \frac{3 t ^{2}}{8}

\int 6 d t = 6 t

= \frac{3 t ^{2}}{8} + 6 t

Füge eine Konstante zur Lösung hinzu

= \frac{3 t ^{2}}{8} + 6 t + C

x \to 3 - lim (9 - x^{2})

t an g e n t f (x) = \frac{x + 2}{x ^{2} - 9}, at x = 1

\int 3 x - 12 d x

\frac{d}{d x} (\frac{3}{2 x} + \frac{4 x}{5})

d er i v a t i v e y = \frac{3 x + 4}{2 x - 1}