integral from 1 to 3 of 7e^{x+2}

Lösung

\int_{1}^{3} 7 e^{x + 2} d x

7 e^{5} - 7 e^{3}

Dezimale

898.29335 \dots

Schritte zur Lösung

\int_{1}^{3} 7 e^{x + 2} d x

Entferne die Konstante:

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

= 7 \cdot \int_{1}^{3} e^{x + 2} d x

e^{x + 2} = e^{2} e^{x}

= 7 \cdot \int_{1}^{3} e^{2} e^{x} d x

Entferne die Konstante:

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

= 7 e^{2} \cdot \int_{1}^{3} e^{x} d x

Nutze das gemeinsame Integral :

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

= 7 e^{2} [e^{x}]_{1}^{3}

Berechne die Grenzen:

e^{3} - e

= 7 e^{2} (e^{3} - e)

Vereinfache

= 7 e^{5} - 7 e^{3}

\int_{1}^{3} \frac{x}{3 x ^{2} + 1} d x

\int_{0}^{2} - 5 x^{2} d x

\int_{0}^{\frac{7}{2}} 7 x - 2 x^{2} d x

\int_{0}^{1} \frac{7}{( 3 x + 4 ) ( 5 x + 2 )} d x

\int_{0}^{8} \frac{1}{x ^{\frac{2}{3}}} d x