integral from 0 to 5 of e^{sqrt(3x+1)}

Lösung

\int_{0}^{5} e^{3 x + 1} d x

2 e^{4}

Dezimale

109.19630 \dots

Schritte zur Lösung

\int_{0}^{5} e^{3 x + 1} d x

Wende U-Substitution an

= \int_{1}^{4} \frac{2 e ^{u} u}{3} d u

Entferne die Konstante:

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

= \frac{2}{3} \cdot \int_{1}^{4} e^{u} u d u

Wende die partielle Integration an

= \frac{2}{3} [e^{u} u - \int e^{u} d u]_{1}^{4}

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

= \frac{2}{3} [e^{u} u - e^{u}]_{1}^{4}

Berechne die Grenzen:

3 e^{4}

= \frac{2}{3} \cdot 3 e^{4}

Vereinfache

= 2 e^{4}

\int_{0}^{\infty} 5 e^{- 5 t} d t

\int_{8}^{6} (\frac{q ^{2}}{4} - 5 q + 25) d q

\int_{1}^{x^{2}} t^{2} d t

\int_{0}^{5} \frac{e ^{x}}{2 + e ^{x}} d x

\int_{0}^{4} \frac{1}{9 x ^{2} + 25} d x