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

Lösung

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

\frac{5 5 - 1}{6}

Dezimale

1.69672 \dots

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende U-Substitution an

= 2 \cdot \int_{1}^{5} \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}^{5} u d u

Wende die Potenzregel an

= 2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Vereinfache

2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5} : \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

Berechne die Grenzen:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{10 5}{3} - \frac{2}{3})

Vereinfache

= \frac{5 5 - 1}{6}

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