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}^{1} x \frac{2 ( x + 1 )}{3} d x

\int_{1}^{1.5} \frac{3}{8} x^{2} d x

\int_{- 2}^{- 1} \frac{3}{y} d y

\int_{0}^{3} 6 x ln (x) d x

\int_{4}^{6} \frac{1}{x - 4} 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