integral from 0 to 1 of 5xe^{3x}

Lösung

\int_{0}^{1} 5 x e^{3 x} d x

\frac{5 ( 2 e ^{3} + 1 )}{9}

Dezimale

22.87281 \dots

Schritte zur Lösung

\int_{0}^{1} 5 x e^{3 x} d x

Entferne die Konstante:

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

= 5 \cdot \int_{0}^{1} x e^{3 x} d x

Wende U-Substitution an

= 5 \cdot \int_{0}^{3} \frac{e ^{u} u}{9} d u

Entferne die Konstante:

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

= 5 \cdot \frac{1}{9} \cdot \int_{0}^{3} e^{u} u d u

Wende die partielle Integration an

= 5 \cdot \frac{1}{9} [e^{u} u - \int e^{u} d u]_{0}^{3}

\int e^{u} d u = e^{u}

= 5 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{3}

Vereinfache

5 \cdot \frac{1}{9} [e^{u} u - e^{u}]_{0}^{3} : \frac{5}{9} [e^{u} u - e^{u}]_{0}^{3}

= \frac{5}{9} [e^{u} u - e^{u}]_{0}^{3}

Berechne die Grenzen:

2 e^{3} + 1

= \frac{5}{9} (2 e^{3} + 1)

Vereinfache

= \frac{5 ( 2 e ^{3} + 1 )}{9}

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