Question

The area (in sq. units) of the region bounded by the curves \( y = 4 \cos x \) and \( y = -|\cos x| \) from \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \) is:

• A. 6
• B. 8
• C. 12
• D. 10

Hint:

When dealing with absolute value functions, break them into pieces based on their sign, and then compute the area by integrating the difference of the functions.

Answer 1

The Correct Option is D

We are given two curves: \[ y = 4 \cos x \quad \text{and} \quad y = -|\cos x| \] The area enclosed between these curves can be found by integrating the difference of the two curves over the interval \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \). First, observe that \( y = -|\cos x| \) will be equivalent to \( y = -\cos x \) in the interval \( x = -\frac{\pi}{2} \) to \( x = \frac{\pi}{2} \), since cosine is non-negative in the given range. The area is given by: \[ \text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[ 4 \cos x - (-\cos x) \right] dx \] \[ \text{Area} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 5 \cos x \, dx \] Now, integrate: \[ \text{Area} = 5 \left[ \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \] \[ \text{Area} = 5 \left[ \sin\left(\frac{\pi}{2}\right) - \sin\left(-\frac{\pi}{2}\right) \right] \] \[ \text{Area} = 5 \left[ 1 - (-1) \right] = 5 \times 2 = 10 \] Thus, the correct answer is option (4), \( \text{Area} = 10 \).