integral from 1 to 9 of (x+1)/(sqrt(x))

Lösung

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

\frac{64}{3}

Dezimale

21.33333 \dots

Schritte zur Lösung

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

Multipliziere aus

\frac{x + 1}{x} : x + \frac{1}{x}

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{1}^{9} x d x = \frac{52}{3}

\int_{1}^{9} \frac{1}{x} d x = 4

= \frac{52}{3} + 4

Vereinfache

\frac{52}{3} + 4 : \frac{64}{3}

= \frac{64}{3}

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