integral from 0 to 2 of 2^{-θ}

Lösung

\int_{0}^{2} 2^{- θ} d θ

\frac{3}{4 ln ( 2 )}

Dezimale

1.08202 \dots

Schritte zur Lösung

\int_{0}^{2} 2^{- θ} d θ

Wende U-Substitution an

= \int_{0}^{- 2} - 2^{u} d u

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

= - \int_{- 2}^{0} - 2^{u} d u

Entferne die Konstante:

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

= - (- \int_{- 2}^{0} 2^{u} d u)

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

= - (- [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0})

Vereinfache

= [\frac{2 ^{u}}{ln ( 2 )}]_{- 2}^{0}

Berechne die Grenzen:

\frac{3}{4 ln ( 2 )}

= \frac{3}{4 ln ( 2 )}

\int sin^{3} (4 x) cos (4 x) d x

\frac{\partial}{\partial x} (x^{- 1} y^{2} + x y^{2} + 5 x y)

\int_{1}^{16} \frac{1}{8 - x} d x

\frac{d y}{d x} + 2 x y = y + 4 x - 2

\frac{d}{d x} ((4 x^{3} + 5)^{\frac{3}{2}})