integral from 4 to 7 of e^{2x+1}

Lösung

\int_{4}^{7} e^{2 x + 1} d x

\frac{e ^{15} - e ^{9}}{2}

Dezimale

1630457.14427 \dots

Schritte zur Lösung

\int_{4}^{7} e^{2 x + 1} d x

Wende U-Substitution an

= \int_{9}^{15} 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_{9}^{15} e^{u} d u

Nutze das gemeinsame Integral :

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

= \frac{1}{2} [e^{u}]_{9}^{15}

Berechne die Grenzen:

e^{15} - e^{9}

= \frac{1}{2} (e^{15} - e^{9})

Vereinfache

= \frac{e ^{15} - e ^{9}}{2}

\int_{2}^{1} (1 + \frac{z}{2}) d z

\int_{1}^{x} sin (x) d x

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

\int_{0}^{2} x e^{\frac{- x ^{2} - 1}{2}} d x

\int_{1}^{4} - x - (2 - x) d x