integral from 0 to 1 of 4xe^{5x}

Lösung

\int_{0}^{1} 4 x e^{5 x} d x

\frac{4 ( 4 e ^{5} + 1 )}{25}

Dezimale

95.14442 \dots

Schritte zur Lösung

\int_{0}^{1} 4 x e^{5 x} d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} x e^{5 x} d x

Wende U-Substitution an

= 4 \cdot \int_{0}^{5} \frac{e ^{u} u}{25} d u

Entferne die Konstante:

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

= 4 \cdot \frac{1}{25} \cdot \int_{0}^{5} e^{u} u d u

Wende die partielle Integration an

= 4 \cdot \frac{1}{25} [e^{u} u - \int e^{u} d u]_{0}^{5}

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

= 4 \cdot \frac{1}{25} [e^{u} u - e^{u}]_{0}^{5}

Vereinfache

4 \cdot \frac{1}{25} [e^{u} u - e^{u}]_{0}^{5} : \frac{4}{25} [e^{u} u - e^{u}]_{0}^{5}

= \frac{4}{25} [e^{u} u - e^{u}]_{0}^{5}

Berechne die Grenzen:

4 e^{5} + 1

= \frac{4}{25} (4 e^{5} + 1)

Vereinfache

= \frac{4 ( 4 e ^{5} + 1 )}{25}

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