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The given problem asks to calculate the integral:

\[
\int_0^2 \sqrt{2x} \, dx
\]

### Solution:

1. First, rewrite the integrand:
   \[
   \sqrt{2x} = \sqrt{2} \cdot \sqrt{x}
   \]
   So the integral becomes:
   \[
   \sqrt{2} \int_0^2 \sqrt{x} \, dx
   \]

2. The integral of \( \sqrt{x} \) can be computed as:
   \[
   \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2}
   \]

3. Now apply the limits of integration:
   \[
   \sqrt{2} \left[ \frac{2}{3} x^{3/2} \right]_0^2
   \]

4. First, substitute \(x = 2\):
   \[
   \frac{2}{3} \cdot 2^{3/2} = \frac{2}{3} \cdot 2 \sqrt{2} = \frac{4\sqrt{2}}{3}
   \]

5. At \(x = 0\), the value is 0.

6. Now multiply by \(\sqrt{2}\):
   \[
   \sqrt{2} \cdot \frac{4\sqrt{2}}{3} = \frac{4 \cdot 2}{3} = \frac{8}{3}
   \]

Thus, the value of the integral is \(\frac{8}{3}\).

The correct answer is **\(\frac{8}{3}\)**.

---

Would you like more details or explanations on any part of the solution? Here are some follow-up questions you might find helpful:

1. How do you solve integrals involving square roots in general?
2. What is the geometric interpretation of definite integrals?
3. Can you apply substitution to simplify similar integrals?
4. How do you handle definite integrals with more complex bounds?
5. How does this integral compare with others involving polynomials?

**Tip**: Always check if a constant factor can be factored out of the integral to simplify your work.

Να υπολογιστεί η τιμή του παρακάτω ολοκληρώματος: ∫₀² √(2x) dx

Learn how to calculate the value of the definite integral ∫₀² √(2x) dx using the Fundamental Theorem of Calculus. This problem covers concepts such as definite integrals and radical functions, suitable for high school or early university-level calculus courses.

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