I am doing multivariable calculus, and specifically double integrals. I am facing difficulties finding the domain of the integal, however i am given the following equations: $$1≤2x+y≤2$$ $$0≤x−2y≤1$$ Through these two equations i am supposed to find the area of integrals for each of the variables i.e $x$ and $y$ I set them as simultaneous inequalities but it doesn't seem to help because i get the boundary for xx to be $$2/5≤x≤1$$ and for $y$ to be $$4/5≤y≤0$$ which is obviously WRONG because how can the lowest boundary for the integral possibly have higher value than that in the higher.
Here is where i have posted my question originally please don't get me wrong, I am thankful to all the lovely & helpful people who tried to help (Henning Makholm, Shardulc, AlphaGo) and for the up-voting, its just i was new to this community and i thought they wanted my email to spam me, so i logged in as a guest.
Anyways, i just don't get how all these people arrived at their solution, is there a known way to solve such problems? Thank you all for everything.
$2x + y \ge 1$ is area above a line. $2x + y \le 2$ is the area below a parallel line. Put them together and you have a ribbon between two parallel lines.
And with the other equation also forms a ribbon that follows a different line. The two ribbons intersect to form a parallelogram.
If you "solve" the system of equations you will need to solve
$2x + y = 1\\ x - 2y = 0$
and
$2x + y = 2\\ x - 2y = 0$
etc. 4 pairs of equations to find 4 vertexes.
But if your goal is to do a double integration, what you really want it to say
$x = 2u + v\\ y = u - 2v\\ 1\le 5u \le 2\\ 0\le 5v \le 1$
$dy\,dx = || (\frac {dx}{du},\frac {dy}{du})\times(\frac {dx}{dv},\frac {dy}{dv})|| du\,dv = 5 \,du\,dv $
$\int_0^{\frac15}\int_{\frac15}^{\frac25} f(u,v) 5 \,du \,dv$