The area under the curve \(y = 2\sqrt{x}\) from \(x = 1\) to \(x = 4\) is given by the definite integral:

\(\int_{1}^{4} 2\sqrt{x} \, dx\)

First, rewrite \(2\sqrt{x}\) as \(2x^{0.5}\).

The integral of \(2x^{0.5}\) is:

\(\frac{2}{1.5} x^{1.5} = \frac{4}{3} x^{1.5}\)

Evaluate this from \(x = 1\) to \(x = 4\):

\(\left[ \frac{4}{3} x^{1.5} \right]_{1}^{4} = \frac{4}{3} (4^{1.5}) - \frac{4}{3} (1^{1.5})\)

Calculate \(4^{1.5} = 8\) and \(1^{1.5} = 1\):

\(\frac{4}{3} (8) - \frac{4}{3} (1) = \frac{32}{3} - \frac{4}{3} = \frac{28}{3}\)