integral from 0 to 2 of x/(sqrt(1+4x))

Lösung

\int_{0}^{2} \frac{x}{1 + 4 x} d x

\frac{5}{6}

Dezimale

0.83333 \dots

Schritte zur Lösung

\int_{0}^{2} \frac{x}{1 + 4 x} d x

Wende U-Substitution an

= \int_{1}^{3} \frac{u ^{2} - 1}{8} d u

Entferne die Konstante:

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

= \frac{1}{8} \cdot \int_{1}^{3} u^{2} - 1 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{1}{8} (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 1 d u)

\int_{1}^{3} u^{2} d u = \frac{26}{3}

\int_{1}^{3} 1 d u = 2

= \frac{1}{8} (\frac{26}{3} - 2)

Vereinfache

\frac{1}{8} (\frac{26}{3} - 2) : \frac{5}{6}

= \frac{5}{6}

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

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