The area enclosed by the curve \(y = \sqrt{x}\), the lines \(x=0\) and \(x=4\), and the x-axis can be found by integrating the function \(y = \sqrt{x}\) with respect to \(x\) from 0 to 4.
Area \(A = \int_{0}^{4} \sqrt{x} \, dx\)
Rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\):
\(A = \int_{0}^{4} x^{\frac{1}{2}} \, dx\)
Integrate \(x^{\frac{1}{2}}\) with respect to \(x\):
\(A = \left[ \frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]_{0}^{4}\)
Simplify:
\(A = \frac{2}{3} \left[ x^{\frac{3}{2}} \right]_{0}^{4}\)
Evaluate the definite integral:
\(A = \frac{2}{3} \left( 4^{\frac{3}{2}} - 0^{\frac{3}{2}} \right)\)
\(A = \frac{2}{3} \left( (4^{\frac{1}{2}})^3 - 0 \right)\)
\(A = \frac{2}{3} \left( 2^3 \right)\)
\(A = \frac{2}{3} \cdot 8\)
\(A = \frac{16}{3}\)
The area of the region is \(\frac{16}{3}\) square units.
Correct Answer: \(\frac{16}{3}\) sq. units
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