1. **State the problem:** We need to find the net area and total area under the curve of the function $f(x) = \sin(2x)$ from $x = -\frac{\pi}{2}$ to $x = \frac{\pi}{2}$. 2. **Recall the formulas:** - The net area is the definite integral of $f(x)$ over the interval: $$\text{Net area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx$$ - The total area is the integral of the absolute value of $f(x)$ over the same interval: $$\text{Total area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin(2x)| \, dx$$ 3. **Calculate the net area:** - Use the integral formula: $$\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C$$ - So, $$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$$ - Evaluate the definite integral: $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x) \, dx = \left[-\frac{1}{2} \cos(2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$$ - Calculate the values: $$-\frac{1}{2} \cos\left(2 \times \frac{\pi}{2}\right) + \frac{1}{2} \cos\left(2 \times -\frac{\pi}{2}\right) = -\frac{1}{2} \cos(\pi) + \frac{1}{2} \cos(-\pi)$$ - Since $\cos(\pi) = \cos(-\pi) = -1$: $$-\frac{1}{2} (-1) + \frac{1}{2} (-1) = \frac{1}{2} - \frac{1}{2} = 0$$ 4. **Calculate the total area:** - The function $\sin(2x)$ is zero at $x=0$ within the interval. - On $[-\frac{\pi}{2},0]$, $\sin(2x) \leq 0$, so $|\sin(2x)| = -\sin(2x)$. - On $[0, \frac{\pi}{2}]$, $\sin(2x) \geq 0$, so $|\sin(2x)| = \sin(2x)$. - Therefore, $$\text{Total area} = \int_{-\frac{\pi}{2}}^{0} -\sin(2x) \, dx + \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx$$ 5. **Evaluate each integral:** - First integral: $$\int -\sin(2x) \, dx = \int -\sin(2x) \, dx = \frac{1}{2} \cos(2x) + C$$ - Evaluate from $-\frac{\pi}{2}$ to $0$: $$\left[\frac{1}{2} \cos(2x)\right]_{-\frac{\pi}{2}}^{0} = \frac{1}{2} \cos(0) - \frac{1}{2} \cos(-\pi) = \frac{1}{2} (1) - \frac{1}{2} (-1) = \frac{1}{2} + \frac{1}{2} = 1$$ - Second integral: $$\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C$$ - Evaluate from $0$ to $\frac{\pi}{2}$: $$\left[-\frac{1}{2} \cos(2x)\right]_{0}^{\frac{\pi}{2}} = -\frac{1}{2} \cos(\pi) + \frac{1}{2} \cos(0) = -\frac{1}{2} (-1) + \frac{1}{2} (1) = \frac{1}{2} + \frac{1}{2} = 1$$ 6. **Sum the two parts for total area:** $$1 + 1 = 2$$ **Final answers:** - Net area = $0$ - Total area = $2$