The area (in square units) bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$ in first quadrant is equal to:

The curve is $y = \cos x$ over the interval $x = 0$ to $x = 2\pi$.

The region is restricted to the first Cartesian quadrant, meaning both $x > 0$ and $y > 0$.

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Over the interval $[0, 2\pi]$, $y = \cos x$ is positive ($y > 0$) in two sections:

$x = 0$ to $x = \pi/2$

$x = 3\pi/2$ to $x = 2\pi$

The section from $x = \pi/2$ to $x = 3\pi/2$ has negative y-values and lies in the fourth quadrant, so it is excluded.

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$\text{Area} = \int_{0}^{\pi/2} \cos x \, dx + \int_{3\pi/2}^{2\pi} \cos x \, dx$

$\text{Area} = [\sin x]_{0}^{\pi/2} + [\sin x]_{3\pi/2}^{2\pi}$

$\text{Area} = (\sin(\pi/2) - \sin(0)) + (\sin(2\pi) - \sin(3\pi/2))$

$\text{Area} = (1 - 0) + (0 - (-1))$

$\text{Area} = 1 + 1$

$\text{Area} = 2$

Therefore, the area bounded in the first quadrant is $\boxed{2}$ square units.

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Common Trap:

Calculating the total bounded area over the entire interval without the first quadrant constraint incorporates the excluded section:

$\text{Total Area} = \int_{0}^{\pi/2} \cos x \, dx + \left| \int_{\pi/2}^{3\pi/2} \cos x \, dx \right| + \int_{3\pi/2}^{2\pi} \cos x \, dx$

$\text{Total Area} = 1 + |-2| + 1$

$\text{Total Area} = 4$

Obtaining an answer of $4$ is a result of ignoring the condition that the area must lie strictly in the first quadrant where $y > 0$.