integral from 0 to 24 of x/(sqrt(9+3x))

Lösung

\int_{0}^{24} \frac{x}{9 + 3 x} d x

40

Schritte zur Lösung

\int_{0}^{24} \frac{x}{9 + 3 x} d x

Wende U-Substitution an

= \int_{3}^{9} \frac{2 ( u ^{2} - 9 )}{9} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{2}{9} \cdot \int_{3}^{9} u^{2} - 9 d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{2}{9} (\int_{3}^{9} u^{2} d u - \int_{3}^{9} 9 d u)

\int_{3}^{9} u^{2} d u = 234

\int_{3}^{9} 9 d u = 54

= \frac{2}{9} (234 - 54)

Vereinfache

\frac{2}{9} (234 - 54) : 40

= 40

t a y l or ln (x + 3), - 1

\frac{d}{d t} ((t^{4} - 2 t)^{2} - 5)

\int_{0}^{9} 81 - x^{2} d x

im pl i c i t x^{y} = y^{x}

\int \frac{x ^{3}}{e ^{x^{4}}} d x