$\int_{0}^{2} \frac{4 x}{\left(4-x^{2}\right)^{\frac{1}{2}}} d x$

Step 1 ：The problem is asking for the definite integral of the function \(\frac{4x}{\sqrt{4-x^2}}\) from 0 to 2. The integral of a function can be thought of as the area under the curve of the function.

Step 2 ：The definite integral of a function can be found by finding the antiderivative of the function and then evaluating it at the upper and lower limits of integration.

Step 3 ：To find the antiderivative of \(\frac{4x}{\sqrt{4-x^2}}\), we can use the substitution method. Let \(u = 4 - x^2\), then \(du = -2x dx\). So, the integral becomes \(-2\int \frac{du}{\sqrt{u}}\)

Step 4 ：The antiderivative of \(\frac{1}{\sqrt{u}}\) is \(2\sqrt{u}\), so the antiderivative of our function is \(-4\sqrt{4 - x^2}\)

Step 5 ：Evaluate the antiderivative at the upper limit of integration (2) and subtract the value of the antiderivative at the lower limit of integration (0).

Step 6 ：The definite integral is \(-4\sqrt{4 - 2^2} - (-4\sqrt{4 - 0^2}) = -4\sqrt{0} + 4\sqrt{4} = 0 + 8 = 8\).

Step 7 ：Final Answer: \(\boxed{8}\)