Question

The area of the shaded region is ... sq.units.

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

Hint:

Split the area at x = 9pi/4 and integrate sin x first, then cos x. The two parts are equal.

Answer 1

The Correct Option is A

Step 1: Understanding the Figure:
The graph shows \(y = \sin x\) rising from the point \((2\pi, 0)\) and \(y = \cos x\) falling to the point \((5\pi/2, 0)\). The two curves meet at \(x = 9\pi/4\). The shaded region lies under both curves, above the X-axis.

Step 2: Key Formula or Approach:
Split the area at the intersection \(x = 9\pi/4\). From \(2\pi\) to \(9\pi/4\) the top boundary is \(\sin x\). From \(9\pi/4\) to \(5\pi/2\) the top boundary is \(\cos x\). \[ A = \int_{2\pi}^{9\pi/4} \sin x\,dx + \int_{9\pi/4}^{5\pi/2} \cos x\,dx \]

Step 3: First part:
\[ \int_{2\pi}^{9\pi/4}\sin x\,dx = \left[-\cos x\right]_{2\pi}^{9\pi/4} = -\cos\frac{9\pi}{4} + \cos 2\pi = 1 - \frac{1}{\sqrt2} \]

Step 4: Second part:
\[ \int_{9\pi/4}^{5\pi/2}\cos x\,dx = \left[\sin x\right]_{9\pi/4}^{5\pi/2} = \sin\frac{5\pi}{2} - \sin\frac{9\pi}{4} = 1 - \frac{1}{\sqrt2} \]

Step 5: Add:
\[ A = 2 - \frac{2}{\sqrt2} = 2 - \sqrt2 \]

Step 6: Why the other options are wrong:
\(2 + \sqrt2\) comes from adding \(\frac{1}{\sqrt2}\) instead of subtracting it. \(\sqrt2\) and 2 do not match either integral. The region is small, and \(2 - \sqrt2 \approx 0.586\) fits.

Final Answer:
The shaded area is \(2 - \sqrt2\) square units, option (A). \[ \boxed{2-\sqrt2} \]