integral from 1 to 2e of (2/x+e^x)

Lösung

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

2 ln (2 e) + e^{2 e} - e

Dezimale

230.31967 \dots

Schritte zur Lösung

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

Wende die Summenregel an:

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

= \int_{1}^{2 e} \frac{2}{x} d x + \int_{1}^{2 e} e^{x} d x

\int_{1}^{2 e} \frac{2}{x} d x = 2 ln (2 e)

\int_{1}^{2 e} e^{x} d x = e^{2 e} - e

= 2 ln (2 e) + e^{2 e} - e

\int_{1}^{x} (t^{2}) d t

\int_{0}^{4} x (4 - x) d x

\int_{0}^{20} ∣ x - 10 ∣ d x

\int_{x^{3}}^{x^{6}} (2 t - 1)^{3} d t

\int_{1}^{2} x (x^{2} - 1)^{2} d x