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

Lösung

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

\frac{2 ( 1 + x ) ^{\frac{3}{2}} - 4 2}{3}

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{2}^{1 + x} u d u

Wende die Potenzregel an

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

Berechne die Grenzen:

\frac{2}{3} (1 + x)^{\frac{3}{2}} - \frac{4 2}{3}

= \frac{2}{3} (1 + x)^{\frac{3}{2}} - \frac{4 2}{3}

Vereinfache

= \frac{2 ( 1 + x ) ^{\frac{3}{2}} - 4 2}{3}

\int_{0}^{1} (x^{\frac{1}{2}} - x^{2}) d x

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