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

Lösung

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

2 (2 - ln (1 + 2))

Dezimale

1.06567 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{1 + 2} \frac{2 ( u - 1 )}{u} d u

Entferne die Konstante:

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

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

Multipliziere aus

\frac{u - 1}{u} : 1 - \frac{1}{u}

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

= 2 (\int_{1}^{1 + 2} 1 d u - \int_{1}^{1 + 2} \frac{1}{u} d u)

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

\int_{1}^{1 + 2} \frac{1}{u} d u = ln (1 + 2)

= 2 (2 - ln (1 + 2))

\int_{0}^{3} (3 x)^{2} - (x^{2})^{2} d x

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