Question

The area bounded by the curves \( y - 1 = \cos x \), \( y = \sin x \) and the X-axis between \( x = 0 \) and \( x = \pi \) is

• A. \( 2 + \frac{\pi}{2} \)
• B. \( -\frac{\pi}{2} \)
• C. \( 2 - \frac{\pi}{2} \)
• D. \( \frac{\pi}{2} \)

Hint:

Sketch the graphs of \( y = 1 + \cos x \) and \( y = \sin x \) in the interval \( [0, \pi] \). Identify the region bounded by these curves and the X-axis. Determine which function is above the X-axis in different parts of the interval. The area will be the integral of the function(s) defining the upper boundary of the region with respect to \( x \) over the given interval.

Answer 1

The Correct Option is D

The curves are \( y = 1 + \cos x \) and \( y = \sin x \).
We need to find the area bounded by these curves and the X-axis between \( x = 0 \) and \( x = \pi \).
First, let's find the points of intersection with the X-axis (\( y = 0 \)).
For \( y = 1 + \cos x \): \( 0 = 1 + \cos x \implies \cos x = -1 \).
In the interval \( [0, \pi] \), this occurs at \( x = \pi \).
For \( y = \sin x \): \( 0 = \sin x \).
In the interval \( [0, \pi] \), this occurs at \( x = 0 \) and \( x = \pi \).
Now, let's find where the two curves intersect: \( 1 + \cos x = \sin x \) \( 1 = \sin x - \cos x \) Multiply and divide by \( \sqrt{2} \): \( 1 = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x \right) \) \( \frac{1}{\sqrt{2}} = \sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} \) \( \frac{1}{\sqrt{2}} = \sin \left( x - \frac{\pi}{4} \right) \) In the interval \( [0, \pi] \), \( -\frac{\pi}{4} \le x - \frac{\pi}{4} \le \frac{3\pi}{4} \).
The values of \( x - \frac{\pi}{4} \) for which \( \sin \left( x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \) are \( \frac{\pi}{4} \).
So, \( x - \frac{\pi}{4} = \frac{\pi}{4} \implies x = \frac{\pi}{2} \).
At \( x = \frac{\pi}{2} \), \( y = \sin \frac{\pi}{2} = 1 \) and \( y = 1 + \cos \frac{\pi}{2} = 1 + 0 = 1 \).
The area is given by the sum of the areas under each curve above the X-axis in the interval \( [0, \pi] \).
Area under \( y = \sin x \) from \( 0 \) to \( \pi \): \( A_1 = \int_0^{\pi} \sin x dx = [-\cos x]_0^{\pi} = -\cos \pi - (-\cos 0) = -(-1) - (-1) = 1 + 1 = 2 \) Area under \( y = 1 + \cos x \) from \( 0 \) to \( \pi \): Since \( 1 + \cos x \ge 0 \) for \( x \in [0, \pi] \), the area is: \( A_2 = \int_0^{\pi} (1 + \cos x) dx = [x + \sin x]_0^{\pi} = (\pi + \sin \pi) - (0 + \sin 0) = (\pi + 0) - (0 + 0) = \pi \) The area bounded by the curves and the X-axis is the area under \( y = \sin x \) since \( \sin x \le 1 + \cos x \) in the interval.
Final Answer: The final answer is $\boxed{\frac{\pi}{2}}$