integral from 4 to 5 of 2^{-x}

Lösung

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

\frac{1}{32 ln ( 2 )}

Dezimale

0.04508 \dots

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{- 4}^{- 5} - 2^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 5}^{- 4} - 2^{u} d u

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{- 5}^{- 4} 2^{u} d u)

\int a^{x} d x = \frac{a ^{x}}{ln a}

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 5}^{- 4})

Vereinfache

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 5}^{- 4}

Berechne die Grenzen:

\frac{1}{32 ln ( 2 )}

= \frac{1}{32 ln ( 2 )}

\int_{- 2}^{0} (8 x^{3} - 7 x^{2}) d x

\int_{- 2}^{2} 3 - ∣ x ∣ d x

\int_{0}^{9} x + 1 d x

\int_{3}^{6} x^{3} + x - 2 d x

\int_{1}^{9} (\frac{4}{x} - e^{- x}) d x