If the area of the region in first quadrant, bounded by the curve y² = 9x, x = 2, x = 4 and the x-axis is a + b\(\sqrt{2}\) then the value of a + b is

12

Given curve, y² = 9x

we have area between x = 2, x = 4

∴ we have to find area between 

x = 2 and x = 4

∴ we have to find area of ABCD.

Area of BCFE = $\int_2^4y.dx$

we know that

y² = 9x

taking square root on both the sides

$y = ±\sqrt{9x}$

$y = ±3\sqrt{x}$

Since ABCD is in 1st quadrant.

we have positive value of y.

$∴y=3\sqrt{x}$

Area of ABCD = $\int_2^4y.dx$

$=3\int_2^4\sqrt{x}dx⇒3\int_2^4(x)^{\frac{1}{2}}dx⇒3\begin{bmatrix}\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{bmatrix}_2^4$

$⇒3\begin{bmatrix}\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\end{bmatrix}_2^4⇒3×\frac{2}{3}\begin{bmatrix}x^{\frac{3}{2}}\end{bmatrix}_2^4⇒2[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}]$

$2[((4)^{\frac{1}{2}})^3-((2)^{\frac{1}{2}})^3]⇒2[(2)^3-(\sqrt{2})^3]⇒2[8-2\sqrt{2}]$

$16-4\sqrt{2}$

Thus, Area = $16-4\sqrt{2}$ square units.

Then value of a + b

= a + b = 12 i.e. 16 - 4 = 12

So, correct option is 2.