integral from 0 to 1 of 2^{2x+1}

Lösung

\int_{0}^{1} 2^{2 x + 1} d x

\frac{3}{ln ( 2 )}

Dezimale

4.32808 \dots

Schritte zur Lösung

\int_{0}^{1} 2^{2 x + 1} d x

Wende U-Substitution an

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

Vereinfache

2^{u - 1} : 2^{u} \cdot 2^{- 1}

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

Entferne die Konstante:

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

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

Wende Exponentenregel an:

a^{- 1} = \frac{1}{a}

= \frac{1}{2} \cdot \int_{1}^{3} 2^{u} d u

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

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

Berechne die Grenzen:

\frac{6}{ln ( 2 )}

= \frac{1}{2} \cdot \frac{6}{ln ( 2 )}

Vereinfache

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