Plot the graph of sinx from 0 to 2π and the required area is shaded

Now observe that the area from 0 to π is above X-axis and the area from π to 2π is below X-axis

The area below X-axis will be negative

Also the areas under sinx from 0 to π and π to 2π are equal in magnitude but they will have opposite sign

Hence if we integrate sinx from 0 to 2π the two areas will cancel out each other as they have opposite signs and we will end up on 0

So either find area under sinx from 0 to π and multiply it by 2 or split the limit 0 to 2π into 0 to π and π to 2π

Here we will split the limit

y = sinx

Integrating from 0 to 2π

Because where c ∈ (a, b)

Here π ∈ (0, 2π)

Also now observe that when x is from π to 2π sinx is negative

Hence for π to 2π sinx will become -sinx

Hence area bounded by sinx from 0 to 2π is 4 unit²