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:

2

The correct answer is Option (2) → 2

Note: The official NTA answer key marks Option 2 (Value: 2). However, strictly adhering to the standard definition of the “first quadrant” in the Cartesian plane gives a value of 1 (as explained below), since the graph of $y=\cos x$ lies in the first quadrant only from $x=0$ to $x=\frac{\pi}{2}$. Hence, the official answer key appears to follow a different interpretation of the question.

Understand the Constraints

1. The curve: $y = \cos x$

2. The interval: From $x = 0$ to $x = 2\pi$

3. The quadrant constraint: The first quadrant of the Cartesian plane strictly requires both $x$ and $y$ coordinates to be non-negative ($x \ge 0$ and $y \ge 0$).

For the graph of $y = \cos x$, the portion lying entirely within the first quadrant of the Cartesian plane occurs only for:  $0 \le x \le \frac{\pi}{2}$

Therefore, the only region that satisfies being bounded by the curve between $x = 0$ and $2\pi$ while staying strictly within the first quadrant coordinates ($x \ge 0, y \ge 0$) is from $x = 0$ to $x = \frac{\pi}{2}$.

The area $A$ is given by the integral of the function over the valid region: $A = \int_{0}^{\frac{\pi}{2}} \cos x \, dx$

The area bounded by the curve in the first quadrant is 1 square unit.