integral from 0 to 2 of 2tsqrt(1+4t^2)

Lösung

\int_{0}^{2} 2 t 1 + 4 t^{2} d t

\frac{17 17 - 1}{6}

Dezimale

11.51546 \dots

Schritte zur Lösung

\int_{0}^{2} 2 t 1 + 4 t^{2} d t

Entferne die Konstante:

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

= 2 \cdot \int_{0}^{2} t 1 + 4 t^{2} d t

Wende U-Substitution an

= 2 \cdot \int_{1}^{17} \frac{u}{8} d u

Entferne die Konstante:

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

= 2 \cdot \frac{1}{8} \cdot \int_{1}^{17} u d u

Wende die Potenzregel an

= 2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Vereinfache

2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17} : \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

Berechne die Grenzen:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{34 17}{3} - \frac{2}{3})

Vereinfache

= \frac{17 17 - 1}{6}

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