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: Use symmetry:
Shift the origin to $x=2\pi$. Then the curves are $y=\sin t$ and $y=\cos t$ for $t$ from 0 to $\pi/2$, meeting at $t=\pi/4$. The shaded area lies under the lower of the two.

Step 2: Reflect one part:
Reflecting about the line $t=\pi/4$ swaps $\sin t$ and $\cos t$. So the area under $\sin t$ from 0 to $\pi/4$ equals the area under $\cos t$ from $\pi/4$ to $\pi/2$. Hence $A = 2\int_0^{\pi/4}\sin t\,dt$.

Step 3: Compute:
$2[-\cos t]_0^{\pi/4} = 2\left(1 - \frac{\sqrt2}{2}\right) = 2-\sqrt2$.

Step 4: Result:
The area is $2-\sqrt2$ square units, option (A).

Final Answer:
Option (A). \[ \boxed{2-\sqrt2} \]