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

Lösung

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

\frac{e ^{4} - 1}{2 e ^{3}}

Dezimale

1.33424 \dots

Schritte zur Lösung

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

Wende U-Substitution an

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

Entferne die Konstante:

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

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

Nutze das gemeinsame Integral :

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

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

Vereinfache

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

Berechne die Grenzen:

e - \frac{1}{e ^{3}}

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

Vereinfache

= \frac{e ^{4} - 1}{2 e ^{3}}

\int_{1}^{2} π (e^{y})^{2} d y

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