integral from 0 to 2 of sqrt(36t^2+9t^4)

Lösung

\int_{0}^{2} 36 t^{2} + 9 t^{4} d t

162 - 8

Dezimale

14.62741 \dots

Schritte zur Lösung

\int_{0}^{2} 36 t^{2} + 9 t^{4} d t

Faktorisiere

36 t^{2} + 9 t^{4} : 3 t^{2} 4 + t^{2}

= \int_{0}^{2} 3 t^{2} 4 + t^{2} d t

Entferne die Konstante:

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

= 3 \cdot \int_{0}^{2} t^{2} 4 + t^{2} d t

t^{2} = (t),

angenommen

t \geq 0

= 3 \cdot \int_{0}^{2} t 4 + t^{2} d t

Wende U-Substitution an

= 3 \cdot \int_{2}^{22} u^{2} d u

Wende die Potenzregel an

= 3 [\frac{u ^{3}}{3}]_{2}^{22}

Berechne die Grenzen:

\frac{16 2}{3} - \frac{8}{3}

= 3 (\frac{16 2}{3} - \frac{8}{3})

Vereinfache

= 162 - 8

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