Competency-Based
Assessment Made Simple.

**Step 1: Break the integral based on the GIF** The greatest integer function (GIF), denoted by $[x]$, has different integer values over different intervals. In the interval $[-\pi/2, \pi/2]$, we have: - For $-\pi/2 \le x < -3$, $[x] = -2$ is not possible as $-\pi/2 \approx -1.57$ - For $-1 \le x < 0$, $[x] = -1$ - For $0 \le x < 1$, $[x] = 0$ - For $1 \le x \le \pi/2$, $[x] = 1$ So, we break the integral into the following intervals: $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x]+4} = \int_{-\pi/2}^{-1} \frac{dx}{[x]+4} + \int_{-1}^{0} \frac{dx}{[x]+4} + \int_{0}^{1} \frac{dx}{[x]+4} + \int_{1}^{\pi/2} \frac{dx}{[x]+4}$

**Step 2: Evaluate each integral** - $\int_{-\pi/2}^{-1} \frac{dx}{[x]+4} = \int_{-\pi/2}^{-1} \frac{dx}{-2+4} = \int_{-\pi/2}^{-1} \frac{dx}{2} = \frac{1}{2} [x]_{-\pi/2}^{-1} = \frac{1}{2} (-1 + \pi/2) = \frac{\pi}{4} - \frac{1}{2}$ - $\int_{-1}^{0} \frac{dx}{[x]+4} = \int_{-1}^{0} \frac{dx}{-1+4} = \int_{-1}^{0} \frac{dx}{3} = \frac{1}{3} [x]_{-1}^{0} = \frac{1}{3} (0 - (-1)) = \frac{1}{3}$ - $\int_{0}^{1} \frac{dx}{[x]+4} = \int_{0}^{1} \frac{dx}{0+4} = \int_{0}^{1} \frac{dx}{4} = \frac{1}{4} [x]_{0}^{1} = \frac{1}{4} (1 - 0) = \frac{1}{4}$ - $\int_{1}^{\pi/2} \frac{dx}{[x]+4} = \int_{1}^{\pi/2} \frac{dx}{1+4} = \int_{1}^{\pi/2} \frac{dx}{5} = \frac{1}{5} [x]_{1}^{\pi/2} = \frac{1}{5} (\pi/2 - 1) = \frac{\pi}{10} - \frac{1}{5}$

**Step 3: Sum the results** $\int_{-\pi/2}^{\pi/2} \frac{dx}{[x]+4} = (\frac{\pi}{4} - \frac{1}{2}) + \frac{1}{3} + \frac{1}{4} + (\frac{\pi}{10} - \frac{1}{5}) = \frac{\pi}{4} + \frac{\pi}{10} - \frac{1}{2} + \frac{1}{3} + \frac{1}{4} - \frac{1}{5}$ $= \frac{5\pi + 2\pi}{20} - \frac{30 - 20 - 15 + 12}{60} = \frac{7\pi}{20} - \frac{7}{60}$