integral from 1 to 2 of 3^{-θ}

Lösung

\int_{1}^{2} 3^{- θ} d θ

\frac{2}{9 ln ( 3 )}

Dezimale

0.20227 \dots

Schritte zur Lösung

\int_{1}^{2} 3^{- θ} d θ

Wende U-Substitution an

= \int_{- 1}^{- 2} - 3^{u} d u

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

= - \int_{- 2}^{- 1} - 3^{u} d u

Entferne die Konstante:

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

= - (- \int_{- 2}^{- 1} 3^{u} d u)

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

= - (- [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{- 1})

Vereinfache

= [\frac{3 ^{u}}{ln ( 3 )}]_{- 2}^{- 1}

Berechne die Grenzen:

\frac{2}{9 ln ( 3 )}

= \frac{2}{9 ln ( 3 )}

\int 5 x^{4} e^{6 x^{5}} d x

d er i v a t i v e ln (cos (ln (7 x)))

x \to 0 lim (sin (8 x))

\frac{\partial}{\partial x} (x^{2} e^{x y^{2}})

\int \frac{2 x ^{2}}{9 - x ^{2}} d x