integral from 2 to 3 of 1/2 e^{-1/2 x}

Lösung

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

\frac{e ^{\frac{3}{2}} - e}{e ^{\frac{5}{2}}}

Dezimale

0.14474 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

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

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

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

Nutze das gemeinsame Integral :

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

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

Vereinfache

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

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

Berechne die Grenzen:

\frac{e ^{\frac{3}{2}} - e}{e ^{\frac{5}{2}}}

= \frac{e ^{\frac{3}{2}} - e}{e ^{\frac{5}{2}}}

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

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