integral from 2 to x of (t^3+8t-5)

Lösung

\int_{2}^{x} (t^{3} + 8 t - 5) d t

4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10

Schritte zur Lösung

\int_{2}^{x} t^{3} + 8 t - 5 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_{2}^{x} t^{3} d t + \int_{2}^{x} 8 t d t - \int_{2}^{x} 5 d t

\int_{2}^{x} t^{3} d t = \frac{x ^{4}}{4} - 4

\int_{2}^{x} 8 t d t = 8 (\frac{x ^{2}}{2} - 2)

\int_{2}^{x} 5 d t = 5 x - 10

= \frac{x ^{4}}{4} - 4 + 8 (\frac{x ^{2}}{2} - 2) - (5 x - 10)

Vereinfache

\frac{x ^{4}}{4} - 4 + 8 (\frac{x ^{2}}{2} - 2) - (5 x - 10) : 4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10

= 4 x^{2} - 5 x + \frac{x ^{4}}{4} - 10

\frac{d}{d x} (a e^{x} cos (x))

\frac{\partial}{\partial x} (\frac{( 3 x - 3 z )}{y + 3 z})

\int 3 cos (x) - 2 sin (x) d x

t an g e n t e^{x} + 1, at x = 0

a re a y = 3 + 2 x, y = \frac{9 + x}{3}