Given that \(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x}\), find y.

A.

\(2x^{\frac{3}{2}} + c\)

B.

\(\frac{2}{3}x^{\frac{3}{2}} + c\)

C.

\(\frac{3}{2}x^{\frac{3}{2}} + c\)

D.

\(\frac{2}{3}x^{2} + c\)

Correct answer is B

\(\frac{\mathrm d y}{\mathrm d x} = \sqrt{x} = x^{\frac{1}{2}}\)

\(y = \int x^{\frac{1}{2}} \mathrm {d} x\)

= \(\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1} + c = \frac{x^{\frac{3}{2}}}{\frac{3}{2}} + c \)

= \(\frac{2}{3}x^{\frac{3}{2}} + c\)