The diagram shows part of the curve $y = 5 + \frac{4}{e^{2x}}$ intersecting the $y$-axis at point $P$.
The tangent to the curve at point $P$ intersects the $x$-axis at $Q$. Find the area of the shaded region.

1. Find the coordinates of point $P$ (the $y$-intercept of the curve).

At the $y$-axis, $x = 0$.
$y = 5 + \frac{4}{e^{0}} = 5 + 4 = 9$
So, $P$ is $(0, 9)$.

2. Find the gradient of the tangent at $P$.

$y = 5 + 4e^{-2x}$
$\frac{dy}{dx} = 4(-2)e^{-2x} = -8e^{-2x}$
At $x = 0$, $\frac{dy}{dx} = -8e^{0} = -8$

3. Find the equation of the tangent at $P$ and the coordinates of $Q$.

Equation of tangent: $y - 9 = -8(x - 0) \implies y = -8x + 9$
At $Q$, the tangent intersects the $x$-axis, so $y = 0$.
$0 = -8x + 9 \implies x = \frac{9}{8}$
So, $Q$ is $(\frac{9}{8}, 0)$.

4. Set up the integral for the area of the shaded region.

The shaded region is bounded by the curve and the tangent from $x = 0$ to $x = \frac{9}{8}$.
Area = $\int_{0}^{\frac{9}{8}} \left( (5 + 4e^{-2x}) - (-8x + 9) \right) dx$
Area = $\int_{0}^{\frac{9}{8}} (4e^{-2x} + 8x - 4) dx$

5. Evaluate the definite integral.

$\int (4e^{-2x} + 8x - 4) dx = \left[ \frac{4e^{-2x}}{-2} + \frac{8x^2}{2} - 4x \right]_{0}^{\frac{9}{8}}$
$= \left[ -2e^{-2x} + 4x^2 - 4x \right]_{0}^{\frac{9}{8}}$
$= \left( -2e^{-2(\frac{9}{8})} + 4\left(\frac{9}{8}\right)^2 - 4\left(\frac{9}{8}\right) \right) - \left( -2e^{0} + 0 - 0 \right)$
$= \left( -2e^{-\frac{9}{4}} + 4\left(\frac{81}{64}\right) - \frac{9}{2} \right) - (-2)$
$= -2e^{-\frac{9}{4}} + \frac{81}{16} - \frac{72}{16} + \frac{32}{16}$
$= \frac{41}{16} - 2e^{-\frac{9}{4}}$