integral from-1 to 2 of e^{u+1}

Lösung

\int_{- 1}^{2} e^{u + 1} d u

e^{3} - 1

Dezimale

19.08553 \dots

Schritte zur Lösung

\int_{- 1}^{2} e^{u + 1} d u

e^{u + 1} = e^{1} e^{u}

= \int_{- 1}^{2} e^{1} e^{u} d u

Entferne die Konstante:

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

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

Vereinfache

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

Nutze das gemeinsame Integral :

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

= e [e^{u}]_{- 1}^{2}

Berechne die Grenzen:

e^{2} - \frac{1}{e}

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

Vereinfache

= e^{3} - 1

\int_{0}^{4} (x^{2} + 4) d x

\int_{2}^{4} π (- x^{2} + 6 x - 8)^{2} d x

\int_{1}^{4} 2^{5 x} d x

\int_{0}^{\infty} x^{4} d x

\int_{2}^{\infty} - 9 x^{- 2} d x