Compute the area between the parabolas $y=x^2$ , $y=\frac{x^2}{2}$ and the straight line $y =2x$.
I know the area is equal to $\int_{0}^2x^2-\frac{x^2}{2}+\int_{2}^42x-\frac{x^2}{2}$.
However I can't get this result unless I graph all the functions.
How do you calculate the area without having to graph the functions ?
Thanks in advance!
The only common point of the parabolas is $(0,0)$, and this point is also in the line. Moreover, $x^2/2<x^2$ for $x>0$.
The line intersects the parabola $x^2$ in $(2,4)$ and the parabola $x^2/2$ in $(4,8)$. Moreover, $2x>x^2$ if $x\in(0,2)$ and $2x\le x^2$ otherwise. Also, $2x>x^2/2$ if $x\in(0,4)$ and $2x\le x^2/2$ otherwise.
So, the only region bounded by the three curves is the "triangle" whose "sides" are:
That is, the region whose area we are to compute is:
$$\{(x,y)\in\Bbb R^2: 0\le x<2,x^2/2<y<x^2\}\cup\{(x,y)\in\Bbb R^2: 2\le x\le4,x^2/2<y<2x\}$$
(Of course, by "triangle" I mean a closed, simple curve piecewise defined in three differentiable pieces, and these pieces are the "sides".)
Remark: I couldn't help "seeing" the graphs in my mind, so I could easily have missed something. Perhaps some expert in formal geometry can catch a gap in my answer. Perhaps Jordan's theorem or something like that is needed.