The problem asks to evaluate the definite integral $\int_{2}^{4} \sqrt{x-2} \, dx$.

1. Problem Description

The problem asks to evaluate the definite integral $\int_{2}^{4} \sqrt{x-2} \, dx$.

2. Solution Steps

To evaluate the definite integral, we first find the indefinite integral of $\sqrt{x-2}$.

Let $u = x-2$. Then $du = dx$. The indefinite integral becomes:

$\int \sqrt{x-2} \, dx = \int \sqrt{u} \, du = \int u^{1/2} \, du$.

Using the power rule for integration, $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$, we have:

$\int u^{1/2} \, du = \frac{u^{(1/2)+1}}{(1/2)+1} + C = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C$.

Substituting $u = x-2$ back into the expression, we get:

$\int \sqrt{x-2} \, dx = \frac{2}{3}(x-2)^{3/2} + C$.

Now, we evaluate the definite integral:

$\int_{2}^{4} \sqrt{x-2} \, dx = \left[ \frac{2}{3}(x-2)^{3/2} \right]_{2}^{4} = \frac{2}{3}(4-2)^{3/2} - \frac{2}{3}(2-2)^{3/2} = \frac{2}{3}(2)^{3/2} - \frac{2}{3}(0)^{3/2} = \frac{2}{3}(2\sqrt{2}) - 0 = \frac{4\sqrt{2}}{3}$.

3. Final Answer

$\frac{4\sqrt{2}}{3}$