integral from 9 to x^3 of 1/(p^2)

Lösung

\int_{9}^{x^{3}} \frac{1}{p ^{2}} d p

- \frac{1}{x ^{3}} + \frac{1}{9}

Schritte zur Lösung

\int_{9}^{x^{3}} \frac{1}{p ^{2}} d p

Wende die Potenzregel an

= [- \frac{1}{p}]_{9}^{x^{3}}

Berechne die Grenzen:

- \frac{1}{x ^{3}} + \frac{1}{9}

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

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