integral from 1 to 2 of (e^{2t}+1/(2t))

Lösung

\int_{1}^{2} (e^{2 t} + \frac{1}{2 t}) d t

\frac{1}{2} (e^{4} - e^{2} + ln (2))

Dezimale

23.95112 \dots

Schritte zur Lösung

\int_{1}^{2} e^{2 t} + \frac{1}{2 t} d t

Wende U-Substitution an

= \int_{2}^{4} (e^{u} + \frac{1}{u}) \frac{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_{2}^{4} e^{u} + \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

= \frac{1}{2} (\int_{2}^{4} e^{u} d u + \int_{2}^{4} \frac{1}{u} d u)

\int_{2}^{4} e^{u} d u = e^{4} - e^{2}

\int_{2}^{4} \frac{1}{u} d u = ln (2)

= \frac{1}{2} (e^{4} - e^{2} + ln (2))

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