What is the integral from 1 to 8 of \sqrt[3]{p} ?

The integral from 1 to 8 of \sqrt[3]{p} is 45/4

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integral from 1 to 8 of \sqrt[3]{p}

\frac{45}{4}

Decimal Notation

11.25

Solution steps

Apply the Power Rule

: [\frac{3}{4} p^{\frac{4}{3}}]_{1}^{8}

= [\frac{3}{4} p^{\frac{4}{3}}]_{1}^{8}

Compute the boundaries:

\frac{45}{4}

= \frac{45}{4}

\int_{0}^{\frac{π}{4}} tan (x^{2}) d x

\int_{- 1}^{1} (u^{7} + 1) (3 u^{2} + 1) d u

\int_{0}^{4} x \cdot (3 - 5 x) d x

\int_{1}^{4} x e^{2 x} d x

\int_{1}^{2} \frac{3}{7} x^{2} d x

What is the integral from 1 to 8 of \sqrt[3]{p} ?
The integral from 1 to 8 of \sqrt[3]{p} is 45/4