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

The integral from 1 to 8 of (\sqrt[3]{x}+1) is 73/4

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

\frac{73}{4}

Decimal Notation

18.25

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

\int_{1}^{8} 1 d x = 7

= \frac{45}{4} + 7

Simplify

\frac{45}{4} + 7 : \frac{73}{4}

= \frac{73}{4}

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What is the integral from 1 to 8 of (\sqrt[3]{x}+1) ?
The integral from 1 to 8 of (\sqrt[3]{x}+1) is 73/4