What is the integral from 0 to 2 of (x+1)^{2/3} ?

The integral from 0 to 2 of (x+1)^{2/3} is (9*3^{2/3}-3)/5

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integral from 0 to 2 of (x+1)^{2/3}

\int_{0}^{2} (x + 1)^{\frac{2}{3}} d x

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

Decimal Notation

3.14415 \dots

Solution steps

\int_{0}^{2} (x + 1)^{\frac{2}{3}} d x

Apply u-substitution

: \int_{1}^{3} u^{\frac{2}{3}} d u

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

Apply the Power Rule

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

= [\frac{3}{5} u^{\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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What is the integral from 0 to 2 of (x+1)^{2/3} ?
The integral from 0 to 2 of (x+1)^{2/3} is (9*3^{2/3}-3)/5