Question 16

From the previous question, $\alpha_1 = 0$, $\alpha_2 = \frac{\pi}{2}$, $\alpha_3 = 2\pi$, and $\alpha_4 = \frac{5\pi}{2}$.

The area $\beta$ is given by:

$\beta = \int_{\alpha_1}^{\alpha_4} |e^{-x} - e^{-x}(\sin x + \cos x)| dx = \int_{0}^{5\pi/2} e^{-x} |1 - (\sin x + \cos x)| dx$

In $[0, \pi/2]$, $\sin x + \cos x \ge 1$. In $[\pi/2, 2\pi]$, $\sin x + \cos x \le 1$. In $[2\pi, 5\pi/2]$, $\sin x + \cos x \ge 1$.

Let $J(x) = \int e^{-x}(\sin x + \cos x - 1) dx = e^{-x}(1 - \cos x)$.

$\beta = [J(x)]_{0}^{\pi/2} + [-J(x)]_{\pi/2}^{2\pi} + [J(x)]_{2\pi}^{5\pi/2}$

$J(0) = 0$

$J(\frac{\pi}{2}) = e^{-\pi/2}$

$J(2\pi) = 0$

$J(\frac{5\pi}{2}) = e^{-5\pi/2}$

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

We need to find the value of:

$-\frac{1}{\pi} \log_e (\beta - 2e^{-\pi/2}) = -\frac{1}{\pi} \log_e (e^{-5\pi/2}) = -\frac{1}{\pi} \left( -\frac{5\pi}{2} \right) = 2.5$.