integral from-5 to 0 of 2^{1-x}

Lösung

\int_{- 5}^{0} 2^{1 - x} d x

\frac{62}{ln ( 2 )}

Dezimale

89.44709 \dots

Schritte zur Lösung

\int_{- 5}^{0} 2^{1 - x} d x

Wende U-Substitution an

= \int_{6}^{1} - 2^{u} d u

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

= - \int_{1}^{6} - 2^{u} d u

Entferne die Konstante:

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

= - (- \int_{1}^{6} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{6})

Vereinfache

= [\frac{2 ^{u}}{ln ( 2 )}]_{1}^{6}

Berechne die Grenzen:

\frac{62}{ln ( 2 )}

= \frac{62}{ln ( 2 )}

t an g e n t y = \frac{6 x}{x ^{2} + 1}

\frac{\partial}{\partial x} (x sin (2 y))

d er i v a t i v e f (x) = (x^{2} + 3) (x^{2} - 4 x)

\int (2 x + 6) e^{- 4 x} d x

\frac{d}{d x} (7 (cos (x^{2}) - 2 x^{2} sin (x^{2})))