integral from 1 to 2 of (e^{3t}-4/t)

Lösung

\int_{1}^{2} (e^{3 t} - \frac{4}{t}) d t

\frac{e ^{6} - e ^{3}}{3} - 4 ln (2)

Dezimale

125.00849 \dots

Schritte zur Lösung

\int_{1}^{2} e^{3 t} - \frac{4}{t} d t

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^{3 t} d t - \int_{1}^{2} \frac{4}{t} d t

\int_{1}^{2} e^{3 t} d t = \frac{e ^{6} - e ^{3}}{3}

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

= \frac{e ^{6} - e ^{3}}{3} - 4 ln (2)

\int_{3}^{5} x e^{x^{2}} d x

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

\int_{0}^{1} t 9 t^{2} + 16 d t

\int_{- 5}^{3} x^{2} + 2 x - 15 d x

\int_{0}^{2} (3 t - 1)^{60} d t