integral from 1 to 4 of (u-9)/(sqrt(u))

Lösung

\int_{1}^{4} \frac{u - 9}{u} d u

- \frac{40}{3}

Dezimale

- 13.33333 \dots

Schritte zur Lösung

\int_{1}^{4} \frac{u - 9}{u} d u

Multipliziere aus

\frac{u - 9}{u} : u - \frac{9}{u}

= \int_{1}^{4} u - \frac{9}{u} d u

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}^{4} u d u - \int_{1}^{4} \frac{9}{u} d u

\int_{1}^{4} u d u = \frac{14}{3}

\int_{1}^{4} \frac{9}{u} d u = 18

= \frac{14}{3} - 18

Vereinfache

\frac{14}{3} - 18 : - \frac{40}{3}

= - \frac{40}{3}

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