Figure 1 shows the curve with equation $$y = rac{2x}{ oot{3x^2 + 4}}$$ The finite region S, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2 - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 8

To evaluate the integral, we apply the method of substitution. Let:

$u = 3x^2 + 4$

Then:

$\frac{du}{dx} = 6x \, \text{or} \, dx = \frac{du}{6x}$.

When ( x = 0 ) then ( u = 4 ) and when ( x = 2 ) then ( u = 16 ).

Replacing x in the integral:

$V = \pi \int_{4}^{16} \frac{4x^2}{u} \cdot \frac{du}{6x} = \frac{2\pi}{3} \int_{4}^{16} \frac{u - 4}{u} \, du$

This leads us to:

$\int \left( 1 - \frac{4}{u} \right) du = u - 4 \ln |u|$.

Evaluating the integral:

$\bigg[ u - 4 \ln |u| \bigg]_{4}^{16} = \left( 16 - 4 \ln 16 \right) - \left( 4 - 4 \ln 4 \right)$

Upon simplification, we find that the volume is:

$V = \frac{8\pi}{3} \ln 4$.

So, the solid's volume is expressed in the form ( k \ln a ), where ( k = \frac{8\pi}{3} ) and ( a = 4 ).