integral from 1 to 5 of (4/x-e^{-x})

Lösung

\int_{1}^{5} (\frac{4}{x} - e^{- x}) d x

4 ln (5) - \frac{1}{e} + \frac{1}{e ^{5}}

Dezimale

6.07661 \dots

Schritte zur Lösung

\int_{1}^{5} \frac{4}{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}^{5} \frac{4}{x} d x - \int_{1}^{5} e^{- x} d x

\int_{1}^{5} \frac{4}{x} d x = 4 ln (5)

\int_{1}^{5} e^{- x} d x = \frac{1}{e} - \frac{1}{e ^{5}}

= 4 ln (5) - (\frac{1}{e} - \frac{1}{e ^{5}})

Vereinfache

= 4 ln (5) - \frac{1}{e} + \frac{1}{e ^{5}}

\int_{0}^{\frac{1}{2}} x \cdot arcsin (x) d x

\int_{0}^{\frac{π}{2}} 1 - cos (2 x) d x

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

\int_{2}^{23} \frac{1}{4 + z ^{2}} d z

\int_{0}^{5} x^{2} - 4 d x