To find the area between the graphs of $y = \sin(x)$ and $y = \cos(x)$ on the interval $[0, 2\pi]$, we need to determine where these functions intersect within the interval, then integrate accordingly.

Step-by-Step Solution:

1. Find the Intersection Points: The points of intersection occur where $\sin(x) = \cos(x)$. This simplifies to: $\tan(x) = 1$ Solving $x = \frac{\pi}{4}$ within the interval $[0, 2\pi]$, we find additional solutions at $x = \frac{5\pi}{4}$ by symmetry.

2. Set Up the Integrals: We split the interval $[0, 2\pi]$ into three parts: $[0, \frac{\pi}{4}]$, $[\frac{\pi}{4}, \frac{5\pi}{4}]$, and $[\frac{5\pi}{4}, 2\pi]$.

   • On $[0, \frac{\pi}{4}]$: $\cos(x) \geq \sin(x)$
   • On $[\frac{\pi}{4}, \frac{5\pi}{4}]$: $\sin(x) \geq \cos(x)$
   • On $[\frac{5\pi}{4}, 2\pi]$: $\cos(x) \geq \sin(x)$

3. Compute Each Integral Separately: The area $A$ between the curves is the sum of the absolute differences between $\sin(x)$ and $\cos(x)$ over these intervals:

   $A = \int_0^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx + \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin(x) - \cos(x)) \, dx + \int_{\frac{5\pi}{4}}^{2\pi} (\cos(x) - \sin(x)) \, dx$

4. Evaluate Each Integral:

   • For $\int_0^{\frac{\pi}{4}} (\cos(x) - \sin(x)) \, dx$: $= \left[ \sin(x) + \cos(x) \right]_0^{\frac{\pi}{4}} = (\sin(\frac{\pi}{4}) + \cos(\frac{\pi}{4})) - (0 + 1) = \sqrt{2} - 1$

   • For $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} (\sin(x) - \cos(x)) \, dx$: $= \left[-\cos(x) - \sin(x) \right]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \left(-\cos(\frac{5\pi}{4}) - \sin(\frac{5\pi}{4})\right) - \left(-\cos(\frac{\pi}{4}) - \sin(\frac{\pi}{4})\right)$ Substituting values, we get: $= (\sqrt{2} + \sqrt{2}) = 2\sqrt{2}$

   • For $\int_{\frac{5\pi}{4}}^{2\pi} (\cos(x) - \sin(x)) \, dx$: [ = \left[ \sin(x) + \cos(x) \right]_{\frac{5\pi}{4}}^{2\pi} = (0 + 1) - (-\sqrt{2}) = \sqrt{2} -