The value of the definite integral I=∫limits_0^2x√(2-x)dx is:

The correct answer is Option (3) → $\frac{16\sqrt{2}}{15}$

Given integral:

$I = \int_{0}^{2} x \sqrt{2 - x} \, dx$

Substitute $t = 2 - x → dt = -dx, x = 2 - t$

Limits: x = 0 → t = 2, x = 2 → t = 0

$I = \int_{t=2}^{0} (2 - t) \sqrt{t} \, (-dt)$

$I = \int_{0}^{2} (2 - t) \sqrt{t} \, dt$

$I = \int_{0}^{2} 2\sqrt{t} \, dt - \int_{0}^{2} t \sqrt{t} \, dt$

$I = 2 \int_{0}^{2} t^{1/2} dt - \int_{0}^{2} t^{3/2} dt$

Compute integrals:

$2 \int t^{1/2} dt = 2 \cdot \frac{2}{3} t^{3/2} = \frac{4}{3} t^{3/2}$

$\int t^{3/2} dt = \frac{2}{5} t^{5/2}$

Evaluate from 0 to 2:

$I = \left[ \frac{4}{3} t^{3/2} - \frac{2}{5} t^{5/2} \right]_{0}^{2}$

At t = 2:

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

$= \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} = \frac{40 - 24}{15} \sqrt{2} = \frac{16\sqrt{2}}{15}$

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