Find the area beween the curves y=x^2 and x+y=2 and the x axis

[karush]

Find the area beween the curves $y=x^2$ and $x+y=2$ and the x axis

First on graphing these the $x-axis$ seem irrelevant in that it is outside the area to find.

[desmos="-10,10,-10,10"]y=x^2;y=-x+2[/desmos]

 

[Jameson]

How would we find the area between each curve and the x-axis individually?

Where do these curves intersect?

 

[karush]

Set the equations equal to each other.
$x=-2$ $x=1$

 

[Jameson]

Set the equations equal to each other.
$x=-2$ $x=1$

Good! Ok, so now to find the area between them we need to find the area of the bigger part and then subtract out the area of of the smaller part. Just like if we had a whole pizza with a piece missing, we could take the area of the whole pizza minus the area of one slice to find the area of the whole pizza with a piece missing.

In this problem, which function has a larger area or put another way, which function is larger than the other over the interval?

 

[karush]

$\int_{-2 }^{1 } (-x^2 - x+2)\,dx$
$\left[-\frac{{x}^{3}}{2 }-\frac{x}{2}+2x \right]^1_{-2}=\frac{9}{2}$

I hope

 

[MarkFL]

I am thinking, given the original problem statement, that you are to find the following area:

View attachment 4411

 

[karush]

For the interval $0\le x\le2$ I get $-\frac{2}{3}$

 

[MarkFL]

Did you set it up as:

\(\displaystyle A=\int_0^1 x^2\,dx+\int_1^2 2-x\,dx\) ?

 

[karush]

No that would be 2/3

 

[MarkFL]

I get a different result, but I can't tell where you and I differ without seeing your work...:D

 

[Jameson]

Yes, you are correct Mark. Sorry for misreading the question karush. I was thinking of this region in light green but Mark is correct. :)

[desmos="-10,10,-10,10"]y<x^2\left\{-2<x<1\right\};x+y<2\left\{-2<x<1\right\};[/desmos]

 

[karush]

No problem, how did you get the shading in demos?