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 $$ =[\square $$

Question

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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 $$ = 2 $$
Net Area: 0 Total Area: 2

Given the function
for between and , find:
The net area
The total area

Upstudy AI Solution

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Solution

1. Net Area

2. Total Area

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Given the function for between and , find: The net area The total area