integral from 0 to 7 of x/(sqrt(4+3x))

Lösung

\int_{0}^{7} \frac{x}{4 + 3 x} d x

6

Schritte zur Lösung

\int_{0}^{7} \frac{x}{4 + 3 x} d x

Wende U-Substitution an

= \int_{2}^{5} \frac{2 ( u ^{2} - 4 )}{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_{2}^{5} u^{2} - 4 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_{2}^{5} u^{2} d u - \int_{2}^{5} 4 d u)

\int_{2}^{5} u^{2} d u = 39

\int_{2}^{5} 4 d u = 12

= \frac{2}{9} (39 - 12)

Vereinfache

\frac{2}{9} (39 - 12) : 6

= 6

\int_{- 4}^{4} (16 x - \frac{x ^{3}}{3}) d x

\int_{1}^{t} x e^{- x^{2}} d x

\int_{2}^{5} \frac{4}{x} d x

\int_{0}^{1} 4 x d x

\int_{2}^{3} (3 x^{2} + 4 x + 8) d x