First, find the points of intersection by equating the line and the curve:

\(2x + 2 = 5x^{\frac{1}{2}}\)

Rearrange to form a quadratic equation:

\(2x - 5x^{\frac{1}{2}} + 2 = 0\)

Let \(u = x^{\frac{1}{2}}\), then \(u^2 = x\). Substitute to get:

\(2u^2 - 5u + 2 = 0\)

Factor the quadratic equation:

\((2u - 1)(u - 2) = 0\)

Thus, \(u = \frac{1}{2}\) or \(u = 2\)

So, \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) or \(x = 2^2 = 4\)

Now, calculate the area between the curves from \(x = \frac{1}{4}\) to \(x = 4\):

Area = \(\int_{\frac{1}{4}}^{4} (5x^{\frac{1}{2}} - (2x + 2)) \, dx\)

Integrate each term separately:

\(\int 5x^{\frac{1}{2}} \, dx = \frac{10}{3}x^{\frac{3}{2}}\)

\(\int (2x + 2) \, dx = x^2 + 2x\)

Evaluate from \(x = \frac{1}{4}\) to \(x = 4\):

\(\left[ \frac{10}{3}x^{\frac{3}{2}} - (x^2 + 2x) \right]_{\frac{1}{4}}^{4}\)

Calculate:

At \(x = 4\): \(\frac{10}{3}(4^{\frac{3}{2}}) - (4^2 + 2 \times 4) = \frac{80}{3} - 24\)

At \(x = \frac{1}{4}\): \(\frac{10}{3}\left(\left(\frac{1}{4}\right)^{\frac{3}{2}}\right) - \left(\left(\frac{1}{4}\right)^2 + 2\left(\frac{1}{4}\right)\right) = \frac{10}{3} \times \frac{1}{8} - \left(\frac{1}{16} + \frac{1}{2}\right)\)

Combine results:

\(\frac{80}{3} - 24 - \left(\frac{10}{24} - \frac{9}{16}\right)\)

Simplify to get the exact area:

\(\frac{45}{16}\)