Sum of area bounded by the curve

[luka perkovic]

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 r*r(x-2sin(x)+½x+sin(2x)/4) from 0 to 2π and we get r*r(2π-2*0+π+0)-0=3r*r*π).Why I can´t just take for -2sin(x) from 0 to 2π is equal to 4 and equally for sin(2x)/4,because that is the area bounded by the curve?

 

[theodoros.mihos]

An integral like $$ \int_a^b f(x)\,dx $$ calculated as the signed area between ##f(x)## and ##x## axis.
If there are conditions that split this integral to partial integrals in ##[a,b]##, then may have different sign.

 

[luka perkovic]

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.

 

[theodoros.mihos]

$$\int_0^{2π} sin(x)\,dx = 2 \int_0^π sin(x)\,dx$$ may used on cases than something change on ##\pi## value. Generally is zero.
Please, place the foul problem.

 

[Samy_A]

$$\int_0^{2π} sin(x)\,dx = 2 \int_0^π sin(x)\,dx$$

That's not correct.
LHS is 0, RHS is 4.

 

[theodoros.mihos]

Yes, is not correct but when we don't explain what we do, some people thinks that may be correct for some reason.
For example, I don't know what is LHS or RHS. Is something like KGB to me.

 

[Samy_A]

Yes, is not correct but when we don't explain what we do, some people thinks that may be correct for some reason.
For example, I don't know what is LHS or RHS. Is something like KGB to me.

Better having problems with LHS or RHS than with the KGB. :)
LHS is shorthand for the left-hand side of an equation. Similarly, RHS is the right-hand side.

 

[luka perkovic]

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.

 

[theodoros.mihos]

What does "area" exactly mean? Area for me is the paper amount that need painting and for this reason I take right hand result. But the integral have sign and is the left hand result.

 

[Samy_A]

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.

It depends on the question.
The integral of the sine function between 0 and 2π is the LHS, with result 0.
But if you are asked to compute the area between the sine function and the x-axis, for x between 0 and 2π (say for a real estate project), then you have to add the "positive" area between 0 and π, and the "negative" area between π and 2π. In this case the RHS is what you want.