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

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

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

\frac{45}{4}

Decimal Notation

11.25

Solution steps

Apply the Power Rule

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

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

Compute the boundaries:

\frac{45}{4}

= \frac{45}{4}

\frac{d}{d x} (2 x^{2} - 2)

f (x) = 4 x^{2}, at x = 9

4 x^{- 1}

- 4 x^{2} + 6.25 x + 8, 2.5 x + 7

\int (sec (x) - 2 x tan (x) - 2 x) d x

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