integral from 0 to 1 of 8e^{2x}

Lösung

\int_{0}^{1} 8 e^{2 x} d x

4 (e^{2} - 1)

Dezimale

25.55622 \dots

Schritte zur Lösung

\int_{0}^{1} 8 e^{2 x} d x

Entferne die Konstante:

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

= 8 \cdot \int_{0}^{1} e^{2 x} d x

Wende U-Substitution an

= 8 \cdot \int_{0}^{2} e^{u} \frac{1}{2} d u

Entferne die Konstante:

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

= 8 \cdot \frac{1}{2} \cdot \int_{0}^{2} e^{u} d u

Nutze das gemeinsame Integral :

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

= 8 \cdot \frac{1}{2} [e^{u}]_{0}^{2}

Vereinfache

8 \cdot \frac{1}{2} [e^{u}]_{0}^{2} : 4 [e^{u}]_{0}^{2}

= 4 [e^{u}]_{0}^{2}

Berechne die Grenzen:

e^{2} - 1

= 4 (e^{2} - 1)

\int_{5}^{7} x^{2} d x

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

\int_{- a}^{a} 2 x a^{2} - x^{2} d x

\int_{0}^{r} a x^{2} + b x + c d x

\int_{- 1}^{3} e^{u + 1} d u