integral from 0 to x of 9te^{3t}

Lösung

\int_{0}^{x} 9 t e^{3 t} d t

3 x e^{3 x} - e^{3 x} + 1

Schritte zur Lösung

\int_{0}^{x} 9 t e^{3 t} d t

Entferne die Konstante:

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

= 9 \cdot \int_{0}^{x} t e^{3 t} d t

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

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

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

Vereinfache

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

= [e^{u} u - e^{u}]_{0}^{3 x}

Berechne die Grenzen:

3 x e^{3 x} - e^{3 x} + 1

= 3 x e^{3 x} - e^{3 x} + 1

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