Given the function \( f(x)=\sin (2 x) \) for \( x \) between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \), find: The net area \( =[?] \) The total area \( =[\square \)

We start with the given function \[ f(x)=\sin(2x) \] for \[ x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \] ### 1. Net Area The net area is defined as \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x)\,dx. \] **Step 1.1: Compute the antiderivative** We have \[ \int \sin(2x)\,dx = -\frac{1}{2}\cos(2x) + C. \] **Step 1.2: Evaluate the definite integral** Substitute the limits: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x)\,dx = \left[-\frac{1}{2}\cos(2x)\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}. \] At \( x=\frac{\pi}{2} \): \[ -\frac{1}{2}\cos\left(2\cdot\frac{\pi}{2}\right) = -\frac{1}{2}\cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2}. \] At \( x=-\frac{\pi}{2} \): \[ -\frac{1}{2}\cos\left(2\cdot\left(-\frac{\pi}{2}\right)\right) = -\frac{1}{2}\cos(-\pi) = -\frac{1}{2}(-1) = \frac{1}{2}. \] Thus, \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(2x)\,dx = \frac{1}{2} - \frac{1}{2} = 0. \] **Net Area:** \( 0 \) ### 2. Total Area The total area is given by the integral of the absolute value: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left|\sin(2x)\right|\,dx. \] Since \(\sin(2x)\) changes sign in the interval, we split the integral where the function is zero. Note that \(\sin(2x)=0\) when \(2x = 0\), which implies \(x=0\). Thus, we write: \[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left|\sin(2x)\right|\,dx = \int_{-\frac{\pi}{2}}^{0} -\sin(2x)\,dx + \int_{0}^{\frac{\pi}{2}} \sin(2x)\,dx. \] **Step 2.1: Evaluate \(\int_{0}^{\frac{\pi}{2}} \sin(2x)\,dx\)** Using the antiderivative: \[ \int \sin(2x)\,dx = -\frac{1}{2}\cos(2x), \] we have: \[ \int_{0}^{\frac{\pi}{2}} \sin(2x)\,dx = \left[-\frac{1}{2}\cos(2x)\right]_0^{\frac{\pi}{2}}. \] At \( x=\frac{\pi}{2} \): \[ -\frac{1}{2}\cos(\pi) = -\frac{1}{2}(-1) = \frac{1}{2}. \] At \( x=0 \): \[ -\frac{1}{2}\cos(0) = -\frac{1}{2}(1) = -\frac{1}{2}. \] Thus, \[ \int_{0}^{\frac{\pi}{2}} \sin(2x)\,dx = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1. \] **Step 2.2: Evaluate \(\int_{-\frac{\pi}{2}}^{0} -\sin(2x)\,dx\)** The antiderivative of \(-\sin(2x)\) is: \[ \int -\sin(2x)\,dx = \frac{1}{2}\cos(2x). \] Thus, \[ \int_{-\frac{\pi}{2}}^{0} -\sin(2x)\,dx = \left[\frac{1}{2}\cos(2x)\right]_{-\frac{\pi}{2}}^{0}. \] At \( x=0 \): \[ \frac{1}{2}\cos(0) = \frac{1}{2}(1) = \frac{1}{2}. \] At \( x=-\frac{\pi}{2} \): \[ \frac{1}{2}\cos\left(2\cdot\left(-\frac{\pi}{2}\right)\right) = \frac{1}{2}\cos(-\pi) = \frac{1}{2}(-1) = -\frac{1}{2}. \] Thus, \[ \int_{-\frac{\pi}{2}}^{0} -\sin(2x)\,dx = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1. \] **Total Area:** \[ 1+1 = 2. \] ### Final Answers - Net Area \( = 0 \) - Total Area \( = 2 \)