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

Using the quadratic formula:

$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 24}}{2} = \frac{-1 \pm 5}{2}$

Thus, we find:

$u = 2 \quad \text{and} \quad u = -3$

Since $u = e^x$ must be positive, we take $u = 2$. Therefore:

$e^x = 2 \Rightarrow x = \ln 2$.

The area $A$ between the two curves from $x = 0$ to $x = \ln 2$ is given by:

$A = \int_0^{\ln 2} ( (6 - e^{-x}) - e^x ) \, dx$

This simplifies to:

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

Calculating this integral gives:

$A = \left[ 6x + e^{-x} - e^x \right]_0^{\ln 2}$

Evaluating at the bounds:

1. At $x = \ln 2$: $6(\ln 2) + e^{-\ln 2} - e^{\ln 2} = 6(\ln 2) + \frac{1}{2} - 2$

2. At $x = 0$: $6(0) + e^{0} - e^{0} = 1 - 1 = 0$

Thus,

$A = 6 \ln 2 + \frac{1}{2} - 2 = 6 \ln 2 - \frac{3}{2}$.

To express the area in the required form, we note that:

$A = 6 \ln 2 - \frac{3}{2} = 6 \ln 4 - 5$

This confirms that the area of the shaded region is indeed $6 \ln 4 - 5$, as required.