Answer: \(\frac94\).

At \(x=2\),

\(y=\sqrt{4(2)+1}=3.\)

So \(A=(2,3)\).

Differentiate the curve:

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac12(4x+1)^{-\frac12}\cdot4=\frac{2}{\sqrt{4x+1}}.\)

At \(x=2\), the gradient is

\(\frac{2}{3}.\)

The tangent at \(A\) is

\(y-3=\frac23(x-2).\)

To find the \(x\)-intercept of the tangent, put \(y=0\):

\(-3=\frac23(x-2).\)

So

\(-9=2x-4,\)

giving

\(x=-\frac52.\)

The area of the triangle under the tangent from \(x=-\frac52\) to \(x=2\) is

\(\frac12\left(2+\frac52\right)(3)=\frac{27}{4}.\)

The curve meets the \(x\)-axis when

\(\sqrt{4x+1}=0,\)

so \(x=-\frac14\).

The area under the curve from \(-\frac14\) to \(2\) is

\(\int_{-\frac14}^{2}\sqrt{4x+1}\,\mathrm{d}x.\)

Since

\(\int \sqrt{4x+1}\,\mathrm{d}x=\frac{(4x+1)^{\frac32}}{6},\)

the area under the curve is

\(\left[\frac{(4x+1)^{\frac32}}{6}\right]_{-\frac14}^{2} =\frac{9^{\frac32}}{6}-0=\frac{27}{6}=\frac92.\)

Therefore the shaded area is

\(\frac{27}{4}-\frac92=\frac{27}{4}-\frac{18}{4}=\frac94.\)