integral from 1 to 4 of e^{bx-b}

Lösung

\int_{1}^{4} e^{b x - b} d x

\frac{e ^{- b} ( e ^{4 b} - e ^{b} )}{b}

Schritte zur Lösung

\int_{1}^{4} e^{b x - b} d x

e^{b x - b} = e^{- b} e^{b x}

= \int_{1}^{4} e^{- b} e^{b x} d x

Entferne die Konstante:

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

= e^{- b} \cdot \int_{1}^{4} e^{b x} d x

Wende U-Substitution an

= e^{- b} \cdot \int_{b}^{b \cdot 4} \frac{e ^{u}}{b} d u

Entferne die Konstante:

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

= e^{- b} \frac{1}{b} \cdot \int_{b}^{b \cdot 4} e^{u} d u

Nutze das gemeinsame Integral :

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

= e^{- b} \frac{1}{b} [e^{u}]_{b}^{b \cdot 4}

Berechne die Grenzen:

e^{b \cdot 4} - e^{b}

= e^{- b} \frac{1}{b} (e^{b \cdot 4} - e^{b})

Vereinfache

= \frac{e ^{- b} ( e ^{4 b} - e ^{b} )}{b}

\int_{1}^{8} 3 x^{2} d x

\int_{0}^{T} (x^{4} - 8 x + 7) d x

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

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\int_{0}^{1} 9 x^{4} + 16 x^{2} d x