What is the integral from 0 to 2 of x^2(8-x^3) ?

The integral from 0 to 2 of x^2(8-x^3) is 32/3

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integral from 0 to 2 of x^2(8-x^3)

\int_{0}^{2} x^{2} (8 - x^{3}) d x

\frac{32}{3}

Decimal

10.66666 \dots

Solution steps

\int_{0}^{2} x^{2} (8 - x^{3}) d x

Apply Integration By Parts

= [\frac{1}{3} x^{3} (8 - x^{3}) - \int - x^{5} d x]_{0}^{2}

\int - x^{5} d x = - \frac{x ^{6}}{6}

= [\frac{1}{3} x^{3} (8 - x^{3}) - (- \frac{x ^{6}}{6})]_{0}^{2}

Simplify

[\frac{1}{3} x^{3} (8 - x^{3}) - (- \frac{x ^{6}}{6})]_{0}^{2} : [\frac{1}{6} (16 x^{3} - x^{6})]_{0}^{2}

= [\frac{1}{6} (16 x^{3} - x^{6})]_{0}^{2}

Compute the boundaries:

\frac{32}{3}

= \frac{32}{3}

What is the integral from 0 to 2 of x^2(8-x^3) ?
The integral from 0 to 2 of x^2(8-x^3) is 32/3