integral from 0 to 8.75 of (22+t)

Lösung

\int_{0}^{8.75} (22 + t) d t

230.78125

Schritte zur Lösung

\int_{0}^{8.75} 22 + t 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_{0}^{8.75} 22 d t + \int_{0}^{8.75} t d t

\int_{0}^{8.75} 22 d t = 192.5

\int_{0}^{8.75} t d t = 38.28125

= 192.5 + 38.28125

Vereinfache

= 230.78125

\int_{1}^{3} \frac{1}{x ^{2}} - \frac{6}{x ^{3}} d x

\frac{d}{d x} (\frac{10}{x ^{2}})

d er i v a t i v e f (x) = x^{2} + 2 x, x = 3, x = 5

d er i v a t i v e f (x) = ln (x^{2} + 3)

\frac{\partial}{\partial x} (\frac{x + 3}{x ^{2} - x - 2})