Question

The area under the curve y = cos x on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is given by :

• A. $\frac{1}{2}$ square units
• B. 2 square units
• C. $\pi$ square units
• D. $\frac{\pi}{4}$ square units

Hint:

The cosine function is symmetric (an even function). Therefore, you can simplify the calculation using symmetry:
$\int_{-\pi/2}^{\pi/2} \cos x \, dx = 2 \int_{0}^{\pi/2} \cos x \, dx = 2 [ \sin x ]_{0}^{\pi/2} = 2(1 - 0) = 2$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The area $A$ bounded by a continuous curve $y = f(x)$ and the $x$-axis from $x = a$ to $x = b$ is given by the definite integral: \[ A = \int_{a}^{b} |f(x)| \, dx \] Detailed Explanation:
Let us analyze the function and the interval:
- The function is $y = \cos x$.
- The interval is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
- Within this interval (the first and fourth quadrants), the value of $\cos x$ is always non-negative ($\cos x \ge 0$).
- Therefore, we can drop the absolute value bars: \[ A = \int_{-\pi/2}^{\pi/2} \cos x \, dx \] Integrating the function: \[ A = \Big[ \sin x \Big]_{-\pi/2}^{\pi/2} \] Substitute the upper and lower limits: \[ A = \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \] Using the identity $\sin(-\theta) = -\sin\theta$: \[ A = 1 - (-1) = 2 \text{ square units} \] Thus, the area under the curve is 2 square units.

Step 2: Final Answer:
The area is 2 square units, which corresponds to Option (B).