What is the integral from 1 to 3 of \sqrt[3]{x^2} ?

The integral from 1 to 3 of \sqrt[3]{x^2} is (9*3^{2/3}-3)/5

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integral from 1 to 3 of \sqrt[3]{x^2}

\frac{9 \cdot 3 ^{\frac{2}{3}} - 3}{5}

Decimal Notation

3.14415 \dots

Solution steps

assuming

x \geq 0

= \int_{1}^{3} x^{\frac{2}{3}} d x

Apply the Power Rule

: [\frac{3}{5} x^{\frac{5}{3}}]_{1}^{3}

= [\frac{3}{5} x^{\frac{5}{3}}]_{1}^{3}

Compute the boundaries:

\frac{9 \cdot 3 ^{\frac{2}{3}}}{5} - \frac{3}{5}

= \frac{9 \cdot 3 ^{\frac{2}{3}}}{5} - \frac{3}{5}

Simplify

= \frac{9 \cdot 3 ^{\frac{2}{3}} - 3}{5}

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\int_{- 1}^{1} x^{4} - 1 d x

\int_{4 x}^{4} e^{- y^{2}} d y

What is the integral from 1 to 3 of \sqrt[3]{x^2} ?
The integral from 1 to 3 of \sqrt[3]{x^2} is (9*3^{2/3}-3)/5