integral from 1 to e of (e^x+2x-1)

Lösung

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

e^{2} - 2 e + e^{e}

Dezimale

17.10675 \dots

Schritte zur Lösung

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

= e^{e} - e + e^{2} - 1 - (e - 1)

Vereinfache

e^{e} - e + e^{2} - 1 - (e - 1) : e^{2} - 2 e + e^{e}

= e^{2} - 2 e + e^{e}

\int_{0}^{5} (3 x + 2) d x

\int_{0}^{\frac{3}{4}} 1 + 4 x d x

\int_{- 3}^{5} 1 d x

\int_{0}^{3} (x + 3) d x

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