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

Lösung

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

\frac{e ^{3} - e}{2}

Dezimale

8.68362 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

= \frac{1}{2} [e^{u}]_{1}^{3}

Berechne die Grenzen:

e^{3} - e

= \frac{1}{2} (e^{3} - e)

Vereinfache

= \frac{e ^{3} - e}{2}

\int_{0}^{3} x 3 - x d x

\int_{0}^{3} x^{2} (1 - x) d x

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