The diagram shows part of the curve $y = \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right)$ and $P(3,\,2.25)$ is a point on the curve.

1. Differentiate the equation of the curve with respect to $x$.

$y = \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right)$
$\frac{dy}{dx} = \frac{1}{2}x - \frac{1}{2} \left( \frac{1}{\frac{x}{3}} \right) \left( \frac{1}{3} \right)$
$\frac{dy}{dx} = \frac{1}{2}x - \frac{1}{2x}$

2. Find the gradient of the tangent at $P(3, 2.25)$.

Substitute $x = 3$ into $\frac{dy}{dx}$:
$\text{Gradient, } m = \frac{1}{2}(3) - \frac{1}{2(3)}$
$m = \frac{3}{2} - \frac{1}{6} = \frac{9}{6} - \frac{1}{6} = \frac{8}{6} = \frac{4}{3}$

3. Find the equation of the tangent at $P$.

Using the point-gradient form $y - y_1 = m(x - x_1)$ with $P(3, 2.25)$:
$y - 2.25 = \frac{4}{3}(x - 3)$
$y - \frac{9}{4} = \frac{4}{3}x - 4$
$y = \frac{4}{3}x - 4 + \frac{9}{4}$
$y = \frac{4}{3}x - \frac{7}{4}$

4. Compare with the given equation and conclude.

Since $y = \frac{4}{3}x - \frac{7}{4}$ is not the same as $y = \frac{4}{3}x - 1$, the given equation is not the equation of the tangent.

1. Apply the product rule to differentiate the first term and differentiate the second term

$\frac{d}{dx}\left[x\ln\left(\frac{x}{3}\right) - x\right] = \left[ (1)\ln\left(\frac{x}{3}\right) + x\left(\frac{1}{\frac{x}{3}}\right)\left(\frac{1}{3}\right) \right] - 1$

2. Simplify the expression

$= \ln\left(\frac{x}{3}\right) + x\left(\frac{3}{x}\right)\left(\frac{1}{3}\right) - 1$
$= \ln\left(\frac{x}{3}\right) + 1 - 1$
$= \ln\left(\frac{x}{3}\right)$

1. Set up the definite integral for the area

$\text{Area} = \int_{1}^{3} \left( \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right) \right) dx$

2. Integrate the expression using the result from part (b)

From (b), $\int \ln\left(\frac{x}{3}\right) dx = x\ln\left(\frac{x}{3}\right) - x$.

$\int \left( \frac{1}{4}x^2 - \frac{1}{2}\ln\left(\frac{x}{3}\right) \right) dx = \frac{x^3}{12} - \frac{1}{2}\left( x\ln\left(\frac{x}{3}\right) - x \right)$

3. Evaluate the definite integral by substituting the limits

$\text{Area} = \left[ \frac{x^3}{12} - \frac{1}{2}\left( x\ln\left(\frac{x}{3}\right) - x \right) \right]_{1}^{3}$

Substitute $x = 3$:
$\frac{3^3}{12} - \frac{1}{2}\left( 3\ln(1) - 3 \right) = \frac{27}{12} - \frac{1}{2}(0 - 3) = \frac{9}{4} + \frac{3}{2} = \frac{15}{4}$

Substitute $x = 1$:
$\frac{1^3}{12} - \frac{1}{2}\left( 1\ln\left(\frac{1}{3}\right) - 1 \right) = \frac{1}{12} - \frac{1}{2}(-\ln 3 - 1) = \frac{1}{12} + \frac{1}{2}\ln 3 + \frac{1}{2} = \frac{7}{12} + \frac{1}{2}\ln 3$

$\text{Area} = \frac{15}{4} - \left( \frac{7}{12} + \frac{1}{2}\ln 3 \right)$
$= \frac{45}{12} - \frac{7}{12} - \frac{1}{2}\ln 3$
$= \frac{38}{12} - \frac{1}{2}\ln 3$
$= \frac{19}{6} - \frac{1}{2}\ln 3$