What is the integral from 8 to 27 of (\sqrt[3]{x}-1/(\sqrt[3]{x)}) ?

The integral from 8 to 27 of (\sqrt[3]{x}-1/(\sqrt[3]{x)}) is 165/4

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integral from 8 to 27 of (\sqrt[3]{x}-1/(\sqrt[3]{x)})

\frac{165}{4}

Decimal

41.25

Solution steps

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{195}{4} - \frac{15}{2}

Simplify

\frac{195}{4} - \frac{15}{2} : \frac{165}{4}

= \frac{165}{4}

What is the integral from 8 to 27 of (\sqrt[3]{x}-1/(\sqrt[3]{x)}) ?
The integral from 8 to 27 of (\sqrt[3]{x}-1/(\sqrt[3]{x)}) is 165/4