Step 1: Writing area as an absolute value integral:
The required area is $\displaystyle A=\int_{0}^{2\pi}|\sin x|\,dx$, because area below the axis must still count as positive.
Step 2: Using the periodicity of |sin x|:
The function $|\sin x|$ repeats every $\pi$ units, unlike $\sin x$ itself which repeats every $2\pi$.
So the interval $[0,2\pi]$ contains exactly two identical periods of $|\sin x|$.
Step 3: Computing the area of one period:
On $[0,\pi]$, $\sin x \ge 0$, so $|\sin x|=\sin x$ there.
\[ \int_{0}^{\pi}|\sin x|\,dx=\int_{0}^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=2 \]
Since there are two such equal periods in $[0,2\pi]$:
\[ A=2\times 2=4 \]
Final Answer:
The area comes out the same using the period of $|\sin x|$ directly.
\[ \boxed{\text{Area}=4 \text{ sq units}} \]