Given curve is $y^2 = x \implies y = \sqrt x$
Required area A is bounded by the curve and the lines x = 1 and x = 4 and the x- axis.
Hence $\begin{align*}A = \int_1^4 \sqrt x\; dx \end{align*}$
This is clearly shown in the fig.
$\begin{align*} \int _1^4 x^{\frac{1}{2}} \; dx \end{align*}$
on integrating we get,
$A = \begin{bmatrix} \frac{x^{\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_1^4$
$\; \; \;= \frac{2}{3} \begin{bmatrix} x^{\frac{3}{2}} \end{bmatrix}_1^4$
on applying limits we get,
$A = \frac{2}{3} \begin{bmatrix} 4^{\frac{3}{2}} - 1^{\frac{3}{2}} \end{bmatrix}$
$\; \; \; = \frac{2}{3} \begin{bmatrix} 2^3 - 1 \end{bmatrix}$
$\; \; \; = \frac{14}{3} sq. units$
Hence the required area =$ \frac{14}{3} sq.units.$