To find the area enclosed by the curve \(y = \sqrt{5x + 4}\), the \(x\)-axis, the \(y\)-axis, and the line \(x = 1\), we need to integrate the function from \(x = 0\) to \(x = 1\).

The integral to evaluate is:

\(\int_0^1 \sqrt{5x + 4} \, dx\)

First, rewrite \(\sqrt{5x + 4}\) as \((5x + 4)^{1/2}\).

Using the power rule for integration, the integral becomes:

\(\int (5x + 4)^{1/2} \, dx = \frac{(5x + 4)^{3/2}}{\frac{3}{2} \cdot 5} = \frac{2}{15} (5x + 4)^{3/2}\)

Evaluate this from \(x = 0\) to \(x = 1\):

\(\left[ \frac{2}{15} (5x + 4)^{3/2} \right]_0^1 = \frac{2}{15} ((5 \cdot 1 + 4)^{3/2} - (5 \cdot 0 + 4)^{3/2})\)

\(= \frac{2}{15} (9^{3/2} - 4^{3/2})\)

\(9^{3/2} = 27\) and \(4^{3/2} = 8\), so:

\(\frac{2}{15} (27 - 8) = \frac{2}{15} \cdot 19 = \frac{38}{15} \approx 2.53\)