integral from 1 to 6 of (9/x-e^{-x})

Lösung

\int_{1}^{6} (\frac{9}{x} - e^{- x}) d x

9 ln (6) - \frac{1}{e} + \frac{1}{e ^{6}}

Dezimale

15.76043 \dots

Schritte zur Lösung

\int_{1}^{6} \frac{9}{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}^{6} \frac{9}{x} d x - \int_{1}^{6} e^{- x} d x

\int_{1}^{6} \frac{9}{x} d x = 9 ln (6)

\int_{1}^{6} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{6}}

= 9 ln (6) - (\frac{1}{e} - \frac{1}{e ^{6}})

Vereinfache

= 9 ln (6) - \frac{1}{e} + \frac{1}{e ^{6}}

\int_{0}^{3} (x^{3} - 4 x) d x

\int_{0}^{3} \frac{3}{x + 4} d x

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

\int_{0}^{5} e^{- x^{2}} d x

\int_{- 3}^{1} (e^{x + 3}) d x