integral from 1 to 8 of sqrt(1+81/16 x)

Lösung

\int_{1}^{8} 1 + \frac{81}{16} x d x

\frac{2656 83 - 97 194}{486 2}

Dezimale

33.24021 \dots

Schritte zur Lösung

\int_{1}^{8} 1 + \frac{81}{16} x d x

Wende U-Substitution an

= \int_{\frac{97}{16}}^{\frac{83}{2}} \frac{16 u}{81} d u

Entferne die Konstante:

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

= \frac{16}{81} \cdot \int_{\frac{97}{16}}^{\frac{83}{2}} u d u

Wende die Potenzregel an

= \frac{16}{81} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{97}{16}}^{\frac{83}{2}}

Berechne die Grenzen:

\frac{83 83}{3 2} - \frac{97 97}{96}

= \frac{16}{81} (\frac{83 83}{3 2} - \frac{97 97}{96})

Vereinfache

= \frac{2656 83 - 97 194}{486 2}

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