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

Lösung

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

1 - \frac{1}{2} ln (3)

Dezimale

0.45069 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{0}^{1} \frac{2 u}{1 + 2 u} d u

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{1} \frac{u}{1 + 2 u} d u

Wende U-Substitution an

= 2 \cdot \int_{1}^{3} \frac{v - 1}{4 v} d v

Entferne die Konstante:

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

= 2 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{v - 1}{v} d v

Multipliziere aus

\frac{v - 1}{v} : 1 - \frac{1}{v}

= 2 \cdot \frac{1}{4} \cdot \int_{1}^{3} 1 - \frac{1}{v} d v

Wende die Summenregel an:

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

= 2 \cdot \frac{1}{4} (\int_{1}^{3} 1 d v - \int_{1}^{3} \frac{1}{v} d v)

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

\int_{1}^{3} \frac{1}{v} d v = ln (3)

= 2 \cdot \frac{1}{4} (2 - ln (3))

Vereinfache

2 \cdot \frac{1}{4} (2 - ln (3)) : 1 - \frac{1}{2} ln (3)

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

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