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

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

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integral from 1 to 4 of \sqrt[3]{x}

Decimal Notation

4.01220 \dots

Solution steps

Apply the Power Rule

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

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

Compute the boundaries:

\int - \frac{3}{x} d x

(x + 3)^{2}

\frac{d}{d y} (\frac{x}{( x ^{2} + y ^{2} )})

n = 1 \sum \infty \frac{( - 1 ) ^{n - 1} n ^{2}}{1 2 ^{n}}

\frac{s + 3}{( s + 3 ) ^{2} + 9}

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