integral from 1 to x of 8/(3t^3)

Lösung

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

- \frac{4}{3 x ^{2}} + \frac{4}{3}

Schritte zur Lösung

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

Entferne die Konstante:

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

= \frac{8}{3} \cdot \int_{1}^{x} \frac{1}{t ^{3}} d t

Wende die Potenzregel an

= \frac{8}{3} [- \frac{1}{2 t ^{2}}]_{1}^{x}

Berechne die Grenzen:

- \frac{1}{2 x ^{2}} + \frac{1}{2}

= \frac{8}{3} (- \frac{1}{2 x ^{2}} + \frac{1}{2})

Vereinfache

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

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