The region enclosed between the curves $y = e^x$, $y = 6 - e^{-rac{x}{2}}$ and the line $x = 0$ is shown shaded in the diagram below - AQA - A-Level Maths Mechanics - Question 15 - 2020 - Paper 1

Let e^{-rac{x}{2}} = t, then we have:

t^2 + 3t - 6 = 0.$$ Using the quadratic formula:

t = rac{-b ext{±} ext{sqrt}(b^2 - 4ac)}{2a},$$

we find:

t = rac{-3 ext{±} ext{sqrt}(3^2 - 4(1)(-6))}{2(1)} = rac{-3 ext{±} ext{sqrt}(9 + 24)}{2} = rac{-3 ext{±} 5}{2}.$$ Thus, the solutions for $t$ are:

t_1 = 1 ext{ and } t_2 = -4.$$
Since t = e^{-rac{x}{2}}, we continue with $t_1 = 1$, giving $x = 0$.

Next, we find the second intersection by substituting $t = 1$ back into e^{-rac{x}{2}}:

ightarrow x = 0.$$ Thus, we need to check the other quadratic solution, which yields: $$t = 1.36 ightarrow e^{-rac{x}{2}} ightarrow x ext{ (numerically calculated around)} = 1.386.$$ The limits of integration for the area between the curves will be from $0$ to this computed value.

The area $A$ between the curves from $x=0$ to $x=1.386$ can be expressed as:

A = ext{Area under } y = 6 - e^{-rac{x}{2}} - ext{Area under } y = e^x.

Thus:

A = \int_0^{1.386} (6 - e^{-rac{x}{2}}) \, dx - \int_0^{1.386} e^x \, dx.

Computing these integrals will lead to the required area.

Calculating:

1. For the first integral:

\int (6 - e^{-rac{x}{2}}) \, dx = 6x + 2e^{-rac{x}{2}} + C,
Evaluating from $0$ to $1.386$:

[6(1.386) + 2e^{-rac{1.386}{2}}] - [0 + 2].

2. For the second integral:

$\int e^x \, dx = e^x + C,$
Evaluating from $0$ to $1.386$:

$[e^{1.386} - e^{0}] = e^{1.386} - 1.$

Thus, combining the results:

A = (6(1.386) + 2e^{-rac{1.386}{2}} - (e^{1.386} - 1)),
Solving these yields:

$6 ext{ln} 4 - 5 = ext{Area},$ which justifies that the area is as required.