integral from 0 to x of sqrt(1+9t)

Lösung

\int_{0}^{x} 1 + 9 t d t

\frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 2}{27}

Schritte zur Lösung

\int_{0}^{x} 1 + 9 t d t

Wende U-Substitution an

= \int_{1}^{1 + 9 x} \frac{u}{9} d u

Entferne die Konstante:

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

= \frac{1}{9} \cdot \int_{1}^{1 + 9 x} u d u

Wende die Potenzregel an

= \frac{1}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{1 + 9 x}

Berechne die Grenzen:

\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{2}{3}

= \frac{1}{9} (\frac{2}{3} (1 + 9 x)^{\frac{3}{2}} - \frac{2}{3})

Vereinfache

= \frac{2 ( 1 + 9 x ) ^{\frac{3}{2}} - 2}{27}

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