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

Lösung

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

2 (2 - ln (3))

Dezimale

1.80277 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Multipliziere aus

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

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

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

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

= 2 (2 - ln (3))

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\int_{- 2}^{2} x^{2} d x