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

Lösung

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

2 (2 - ln (3))

Dezimale

1.80277 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Multipliziere aus

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

= 2 \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 (\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 (2 - ln (3))

\int_{3}^{x} \frac{sin ( t )}{t} d t

\int_{- 1}^{1} - 2 x^{2} + 2 d x

\int_{0}^{1} sin (x^{4}) d x

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

\int_{x}^{x^{2}} ln (x) d x