integral from 0 to 3 of (xe^{2x})

Lösung

\int_{0}^{3} (x e^{2 x}) d x

\frac{5 e ^{6} + 1}{4}

Dezimale

504.53599 \dots

Schritte zur Lösung

\int_{0}^{3} x e^{2 x} d x

Wende U-Substitution an

= \int_{0}^{6} \frac{e ^{u} u}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{0}^{6} e^{u} u d u

Wende die partielle Integration an

= \frac{1}{4} [e^{u} u - \int e^{u} d u]_{0}^{6}

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

= \frac{1}{4} [e^{u} u - e^{u}]_{0}^{6}

Berechne die Grenzen:

5 e^{6} + 1

= \frac{1}{4} (5 e^{6} + 1)

Vereinfache

= \frac{5 e ^{6} + 1}{4}

\int_{- 3}^{3} (x^{7} + x^{2} + 2 x + 1) d x

\int_{0}^{3} \frac{1}{x ^{1.7}} d x

\int_{- 2}^{4} cos (x) d x

\int_{0}^{40} 1 - e^{- \frac{x}{22}} d x

\int_{0}^{0.5} sin^{2} (x) d x