The area (in sq.units) of region bounded by $y^2 = 9x$, $x = 2$, $x = 4$ and the $x$-axis in the first quadrant is

The region is bounded by:

• Curve: $y^2 = 9x$ (a parabola)
• Vertical lines: $x = 2$ and $x = 4$
• The $x$-axis (where $y = 0$)
• First quadrant only

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Given $y^2 = 9x$

$y = \pm\sqrt{9x}$

$y = \pm 3\sqrt{x}$

In the first quadrant, $y > 0$, so:

$y = 3\sqrt{x}$

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The area is given by:

$\text{Area} = \int_2^4 y \, dx$

$= \int_2^4 3\sqrt{x} \, dx$

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$A = 3\int_2^4 x^{1/2} \, dx$

$= 3 \left[\frac{x^{3/2}}{3/2}\right]_2^4$

$= 3 \times \frac{2}{3} \left[x^{3/2}\right]_2^4$

$= 2\left[x^{3/2}\right]_2^4$

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$A = 2\left[4^{3/2} - 2^{3/2}\right]$

$4^{3/2} = (4^{1/2})^3 = 2^3 = 8$

$2^{3/2} = (2^{1/2})^3 = (\sqrt{2})^3 = 2\sqrt{2}$

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$A = 2[8 - 2\sqrt{2}]$

$= 2 \times 2[4 - \sqrt{2}]$

$= 4[4 - \sqrt{2}]$

Therefore, the area is $4[4 - \sqrt{2}]$ square units.