What is the integral from 1 to 8 of 1/(\sqrt[3]{x)} ?

The integral from 1 to 8 of 1/(\sqrt[3]{x)} is 9/2

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integral from 1 to 8 of 1/(\sqrt[3]{x)}

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

\frac{9}{2}

Decimal

4.5

Solution steps

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

Apply the Power Rule

: [\frac{3}{2} x^{\frac{2}{3}}]_{1}^{8}

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

Compute the boundaries:

\frac{9}{2}

= \frac{9}{2}

\int_{1}^{5} \frac{12 ( ln ( x ) ) ^{5}}{x} d x

\int_{8}^{17} x x^{2} - 64 d x

\int_{0}^{3} \frac{x e ^{x}}{( 1 + x ) ^{2}} d x

\int_{1}^{e^{1}} ln (x) d x

\int_{- 5}^{- 3} \frac{x}{x ^{2} - 9} d x

What is the integral from 1 to 8 of 1/(\sqrt[3]{x)} ?
The integral from 1 to 8 of 1/(\sqrt[3]{x)} is 9/2