The area of the region bounded by y² = 9x, x = 2, x = 4 and the x-axis in the first quadrant, is

The region is bounded by $y^2 = 9x$, $x = 2$, $x = 4$, and the $x$-axis in the first quadrant.

Given: $y^2 = 9x$

In the first quadrant, $y$ is positive:

$y = \sqrt{9x}$

$y = 3\sqrt{x}$

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The area under the curve from $x = 2$ to $x = 4$ is:

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

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

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$\int 3x^{1/2} \, dx = 3 \times \dfrac{x^{3/2}}{3/2}$

$= 3 \times \dfrac{2}{3} \times x^{3/2}$

$= 2x^{3/2}$

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

$= 2(4)^{3/2} - 2(2)^{3/2}$

For $x = 4$:

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

$2(4)^{3/2} = 2 \times 8 = 16$

For $x = 2$:

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

$2(2)^{3/2} = 2 \times 2\sqrt{2} = 4\sqrt{2}$

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$\text{Area} = 16 - 4\sqrt{2}$ square units

Therefore, the area of the region is $16 - 4\sqrt{2}$ square units.