Passage: Consider the curve C_1 given by y = e^ -x for x ∈ [0, 10π]...

To find the points of intersection of the curves $C_1$ and $C_2$, we equate their equations:

$e^{-x} = e^{-x}(\sin x + \cos x)$

Since $e^{-x} \neq 0$, we have:

$\sin x + \cos x = 1$

$\Rightarrow \sqrt{2}\sin\left(x + \dfrac{\pi}{4}\right) = 1 \Rightarrow \sin\left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{\sqrt{2}}$

The general solution is $x + \dfrac{\pi}{4} = 2k\pi + \dfrac{\pi}{4}$ or $x + \dfrac{\pi}{4} = 2k\pi + \dfrac{3\pi}{4}$ for any integer $k$.

Thus, $x = 2k\pi$ or $x = 2k\pi + \dfrac{\pi}{2}$.

For $x \in [0, 10\pi]$, the first four points of intersection are:

$\alpha_1 = 0$

$\alpha_2 = \dfrac{\pi}{2}$

$\alpha_3 = 2\pi$

$\alpha_4 = 2\pi + \dfrac{\pi}{2} = \dfrac{5\pi}{2}$

The area $\beta$ enclosed between the curves from $x = \alpha_1$ to $x = \alpha_4$ is given by:

$\beta = \int_{0}^{\dfrac{5\pi}{2}} |e^{-x}(\sin x + \cos x - 1)| dx$

We analyze the sign of $\sin x + \cos x - 1$ in the intervals $(0, \dfrac{\pi}{2})$, $(\dfrac{\pi}{2}, 2\pi)$, and $(2\pi, \dfrac{5\pi}{2})$:

For $x \in (0, \dfrac{\pi}{2})$, $\sin x + \cos x > 1$

For $x \in (\dfrac{\pi}{2}, 2\pi)$, $\sin x + \cos x < 1$

For $x \in (2\pi, \dfrac{5\pi}{2})$, $\sin x + \cos x > 1$

Let $I = \int e^{-x}(\sin x + \cos x - 1) dx$. Using integration by parts or standard formulas, we get:

$\int e^{-x}\sin x dx = -\dfrac{e^{-x}}{2}(\sin x + \cos x)$

$\int e^{-x}\cos x dx = \dfrac{e^{-x}}{2}(\sin x - \cos x)$

$\Rightarrow \int e^{-x}(\sin x + \cos x) dx = -e^{-x}\cos x$

Thus, $I = -e^{-x}\cos x + e^{-x} = e^{-x}(1 - \cos x)$.

Now, we evaluate the area $\beta$ by splitting the integral:

$\beta = \int_{0}^{\dfrac{\pi}{2}} e^{-x}(\sin x + \cos x - 1) dx - \int_{\dfrac{\pi}{2}}^{2\pi} e^{-x}(\sin x + \cos x - 1) dx + \int_{2\pi}^{\dfrac{5\pi}{2}} e^{-x}(\sin x + \cos x - 1) dx$

$\beta = \left[ e^{-x}(1 - \cos x) \right]_{0}^{\dfrac{\pi}{2}} - \left[ e^{-x}(1 - \cos x) \right]_{\dfrac{\pi}{2}}^{2\pi} + \left[ e^{-x}(1 - \cos x) \right]_{2\pi}^{\dfrac{5\pi}{2}}$

$\beta = (e^{-\dfrac{\pi}{2}} - 0) - (0 - e^{-\dfrac{\pi}{2}}) + (e^{-\dfrac{5\pi}{2}} - 0)$

$\beta = 2e^{-\dfrac{\pi}{2}} + e^{-\dfrac{5\pi}{2}}$

We are required to find the value of $-\dfrac{1}{\pi}\log_e\left(\beta - 2 e^{-\dfrac{\pi}{2}}\right)$:

$\beta - 2 e^{-\dfrac{\pi}{2}} = e^{-\dfrac{5\pi}{2}}$

$\Rightarrow -\dfrac{1}{\pi}\log_e\left(e^{-\dfrac{5\pi}{2}}\right) = -\dfrac{1}{\pi} \times \left(-\dfrac{5\pi}{2}\right) = \dfrac{5}{2} = 2.5$

Answer: $2.5$