That´s my problem, why we in complex integrals with sin(x) (from 0 to 2 π) use LHS for sin(x)?But when we calculate area bounded only with sinusoid (by 0 to 2π) ,we use RHS.
Is then area of
\int_0^2π sin(x)\,dx
0 or 4?If we take
\int_0^2π sin(x)\,dx = 2 \int_0^π sin(x)\,dx
then area be 4(what is correct because that is area bounded by the curve).I can´t understand how area in this case could be 0 because area is not a vector.Thanks.
Why we sometimes take the area bounded by the curve is sum of positive area and absolute of negative area(e.g. ∫\int_0^2π sin(x)\, dx is equal to 4 or area of ellipse )?But sometimes we just sum positive and negative areas which is equal to 0(e.g. area of cycloid →when we integrate we get...