The diagram shows part of the curve $y = 4 - e^{-2x}$ which crosses the axes at $A$ and at $B$.

1. Find the coordinates of A by setting y = 0

At $A$, the curve crosses the $x$-axis, so $y = 0$.
$4 - e^{-2x} = 0$
$e^{-2x} = 4$
$-2x = \ln 4$
$x = -\frac{1}{2}\ln 4$
So, $A$ is $\left(-\frac{1}{2}\ln 4, 0\right)$.

2. Find the coordinates of B by setting x = 0

At $B$, the curve crosses the $y$-axis, so $x = 0$.
$y = 4 - e^{-2(0)}$
$y = 4 - e^0$
$y = 4 - 1 = 3$
So, $B$ is $(0, 3)$.

1. State the coordinates of $B$ from part (i)

At the $y$-axis, $x = 0$.
$y = 4 - e^{0} = 4 - 1 = 3$
So, $B$ is $(0, 3)$.

2. Find the derivative of the curve to determine the gradient function

$y = 4 - e^{-2x}$
$\frac{dy}{dx} = -(-2)e^{-2x} = 2e^{-2x}$

3. Find the gradient of the tangent and the normal at $B$

At $x = 0$, gradient of tangent $m_t = 2e^{0} = 2$.
Gradient of normal $m_n = -\frac{1}{m_t} = -\frac{1}{2}$.

4. Find the equation of the normal at $B$

Equation of the normal:
$y - 3 = -\frac{1}{2}(x - 0)$
$y = -\frac{1}{2}x + 3$

5. Find the coordinates of $C$ by setting $y = 0$

The normal meets the $x$-axis at $C$, so $y = 0$.
$0 = -\frac{1}{2}x + 3$
$\frac{1}{2}x = 3$
$x = 6$
Coordinates of $C$ are $(6, 0)$.

1. Recall or calculate the coordinates of $A$, $B$, and $C$ from earlier parts

From (i), $A$ is the $x$-intercept of the curve: $4 - e^{-2x} = 0 \implies e^{-2x} = 4 \implies x = -\frac{1}{2}\ln 4 = -\ln 2$. So $A(-\ln 2, 0)$.
$B$ is the $y$-intercept: $x = 0 \implies y = 4 - e^0 = 3$. So $B(0, 3)$.
From (ii), the normal at $B$ has gradient $-\frac{1}{2}$ and equation $y = -\frac{1}{2}x + 3$. Its $x$-intercept is $C(6, 0)$.

2. Set up and evaluate the definite integral for the area under the curve from $A$ to the $y$-axis

$\text{Area}_1 = \int_{-\ln 2}^{0} (4 - e^{-2x}) \, dx$
$= \left[ 4x + \frac{1}{2}e^{-2x} \right]_{-\ln 2}^{0}$

3. Substitute the limits to find the exact area under the curve

$= \left( 0 + \frac{1}{2}e^0 \right) - \left( 4(-\ln 2) + \frac{1}{2}e^{-2(-\ln 2)} \right)$
$= \frac{1}{2} - \left( -4\ln 2 + \frac{1}{2}e^{\ln 4} \right)$
$= \frac{1}{2} - \left( -4\ln 2 + \frac{1}{2}(4) \right)$
$= \frac{1}{2} + 4\ln 2 - 2 = 4\ln 2 - \frac{3}{2}$

4. Calculate the area of the triangle formed by the normal and the axes, and find the total area

The region from $x=0$ to $x=6$ is a right-angled triangle $OBC$.
$\text{Area}_2 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3 = 9$
$\text{Total Area} = \text{Area}_1 + \text{Area}_2 = \left(4\ln 2 - \frac{3}{2}\right) + 9 = 4\ln 2 + \frac{15}{2}$
$\approx 10.3 \text{ units}^2$ (to 3 s.f.)