Question

Find the area of the region bounded by the curve \(y=\sin x\), \(x=0\) and \(x=2\pi\).

Hint:

sin x is positive on [0,π] and negative on [π,2π]; add the absolute areas of both parts.

Answer 1

Step 1: Note the symmetry instead of splitting by sign region-by-region:
$\sin x$ on $[0,2\pi]$ has a hump above the axis on $[0,\pi]$ and an identical-shaped hump below the axis on $[\pi,2\pi]$ (it is just the mirror image, shifted by $\pi$).

Step 2: Find the area of one hump:
$\int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = 1-(-1) = 2$ square units for the first hump.

Step 3: Double it by symmetry:
Because the second hump is congruent in shape (just flipped below the axis), its area is also 2 square units, so total area $= 2\times 2 = 4$.

Final Answer:
Total enclosed area is 4 square units. \[ \boxed{4 \text{ sq. units}} \]