integral from 1 to 2 of tsqrt(9+16t^2)

Lösung

\int_{1}^{2} t 9 + 16 t^{2} d t

\frac{73 73 - 125}{48}

Dezimale

10.38983 \dots

Schritte zur Lösung

\int_{1}^{2} t 9 + 16 t^{2} d t

Wende U-Substitution an

= \int_{25}^{73} \frac{u}{32} d u

Entferne die Konstante:

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

= \frac{1}{32} \cdot \int_{25}^{73} u d u

Wende die Potenzregel an

= \frac{1}{32} [\frac{2}{3} u^{\frac{3}{2}}]_{25}^{73}

Berechne die Grenzen:

\frac{146 73}{3} - \frac{250}{3}

= \frac{1}{32} (\frac{146 73}{3} - \frac{250}{3})

Vereinfache

= \frac{73 73 - 125}{48}

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