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: $y^2 = 9x$y2=9x (a parabola)
• Vertical lines: $x = 2$x=2 and $x = 4$x=4
• The $x$x-axis (where $y = 0$y=0)
• First quadrant only

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Given $y^2 = 9x$y2=9x

$$y = \pm\sqrt{9x}$$y=±9x​

$$y = \pm 3\sqrt{x}$$y=±3x​

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

$$y = 3\sqrt{x}$$y=3x​

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The area is given by:

$$\text{Area} = \int_2^4 y \, dx$$Area=∫24​ydx

$$= \int_2^4 3\sqrt{x} \, dx$$=∫24​3x​dx

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$$A = 3\int_2^4 x^{1/2} \, dx$$A=3∫24​x1/2dx

$$= 3 \left[\frac{x^{3/2}}{3/2}\right]_2^4$$=3[3/2x3/2​]24​

$$= 3 \times \frac{2}{3} \left[x^{3/2}\right]_2^4$$=3×32​[x3/2]24​

$$= 2\left[x^{3/2}\right]_2^4$$=2[x3/2]24​

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$$A = 2\left[4^{3/2} - 2^{3/2}\right]$$A=2[43/2−23/2]

$$4^{3/2} = (4^{1/2})^3 = 2^3 = 8$$43/2=(41/2)3=23=8

$$2^{3/2} = (2^{1/2})^3 = (\sqrt{2})^3 = 2\sqrt{2}$$23/2=(21/2)3=(2​)3=22​

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$$A = 2[8 - 2\sqrt{2}]$$A=2[8−22​]

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

$$= 4[4 - \sqrt{2}]$$=4[4−2​]

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