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# Find the area of the region bounded by the curve $y^2 = x$ and the lines $x = 1, x = 4$ and the $x- axis.$

$(A)\; \frac{14}{3} sq.units.$
$(B)\; \frac{7}{3} sq.units.$
$(C)\; \frac{28}{3} sq.units.$
$(D)\; \frac{11}{3} sq.units.$

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A)
Toolbox:
• To find the area bounded by the curve $y = f(x), x-axis$ and the ordinates $x = a$ and $x = b,$ then the required area is given by.
• \begin{align*} A = \int_a^b y \; {dx} = \int_a^b f(x)\; dx \end{align*}
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 =$\frac{14}{3} sq.units.$