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

Lösung

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

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

Dezimale

5.36392 \dots

Schritte zur Lösung

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

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

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

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

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

\int_{0}^{x} t e^{t^{2}} d t

\int_{0}^{1} cos (x + y) d x

\int_{1}^{3} \frac{x ^{4} + 1}{x ^{2}} d x

\int_{0}^{6} ∣ 3 - x ∣ d x