The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

The value of the definite integral $I = \int_0^2 x\sqrt{2-x} dx$ is:

Let $u = 2 - x$

Then:

$x = 2 - u$

$dx = -du$

Changing the limits:

When $x = 0$: $u = 2$

When $x = 2$: $u = 0$

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$I = \int_0^2 x\sqrt{2-x} \, dx$

$= \int_2^0 (2-u)\sqrt{u} \, (-du)$

$= \int_0^2 (2-u)\sqrt{u} \, du$

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$I = \int_0^2 (2\sqrt{u} - u\sqrt{u}) \, du$

$= \int_0^2 (2u^{1/2} - u^{3/2}) \, du$

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$I = \left[2 \cdot \frac{u^{3/2}}{3/2} - \frac{u^{5/2}}{5/2}\right]_0^2$

$= \left[\frac{4u^{3/2}}{3} - \frac{2u^{5/2}}{5}\right]_0^2$

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Substituting $u = 2$:

$I = \frac{4(2)^{3/2}}{3} - \frac{2(2)^{5/2}}{5}$

Since $(2)^{3/2} = 2\sqrt{2}$ and $(2)^{5/2} = 4\sqrt{2}$:

$= \frac{4 \cdot 2\sqrt{2}}{3} - \frac{2 \cdot 4\sqrt{2}}{5}$

$= \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5}$

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$I = 8\sqrt{2}\left(\frac{1}{3} - \frac{1}{5}\right)$

$= 8\sqrt{2}\left(\frac{5 - 3}{15}\right)$

$= 8\sqrt{2} \cdot \frac{2}{15}$

$= \frac{16\sqrt{2}}{15}$

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Therefore, $I = \dfrac{16\sqrt{2}}{15}$