Find the total area between the $$ x $$-axis and the function on the given interval. $$ f(x)=\sin x $$ $$ [0,2 \pi] $$ $$ [?] $$

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1. Identify the zeros of the function on the interval:
   - The function is $$ f(x)=\sin x $$.
   - It crosses the $$ x $$-axis when $$\sin x=0$$, i.e., at $$ x=0, \pi, 2\pi $$.

2. Split the interval $$[0, 2\pi]$$ into segments where $$ \sin x $$ maintains a constant sign:
   - On $$[0, \pi]$$, $$\sin x \ge 0$$.
   - On $$[\pi, 2\pi]$$, $$\sin x \le 0$$.

3. Calculate the area on $$[0, \pi]$$:
   - The area is given by:
     $$
     A_1 = \int_{0}^{\pi} \sin x \, dx
     $$
   - The antiderivative of $$\sin x$$ is $$-\cos x$$. Thus:
     $$
     A_1 = \left[-\cos x\right]_{0}^{\pi} = \left(-\cos \pi\right) - \left(-\cos 0\right)
     $$
   - Since $$\cos \pi = -1$$ and $$\cos 0 = 1$$, we have:
     $$
     A_1 = (-(-1)) - ( -1 ) = 1 + 1 = 2
     $$

4. Calculate the area on $$[\pi, 2\pi]$$:
   - Since $$\sin x$$ is negative, the area is:
     $$
     A_2 = \int_{\pi}^{2\pi} \left(-\sin x\right) \, dx
     $$
   - The antiderivative of $$-\sin x$$ is $$\cos x$$. Thus:
     $$
     A_2 = \left[\cos x\right]_{\pi}^{2\pi} = \cos (2\pi) - \cos (\pi)
     $$
   - Substituting the values $$\cos 2\pi = 1$$ and $$\cos \pi = -1$$:
     $$
     A_2 = 1 - (-1) = 2
     $$

5. The total area is the sum of the areas on each segment:
   $$
   \text{Total Area} = A_1 + A_2 = 2 + 2 = 4
   $$
The total area between the $$ x $$-axis and $$ f(x) = \sin x $$ on the interval $$[0, 2\pi]$$ is 4 square units.

Find the total area between the
-axis and the function on the
given interval.

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Find the total area between the -axis and the function on the given interval.