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

Lösung

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

\frac{10}{3}

Dezimale

3.33333 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Vereinfache

\frac{1}{2} (\frac{26}{3} - 2) : \frac{10}{3}

= \frac{10}{3}

\int_{0}^{21} - 117 x + 23273 d x

\int_{- π}^{5 π} cos^{2} (x) d x

\int_{0}^{- \infty} e^{- x^{2}} d x

\int_{1}^{4} x^{2} - x - 6 d x

\int_{2}^{5} (x^{2} + \frac{1}{x}) d x