Region bounded by two curves, one is x= and the other is y=

[NYC]

I'm fairly comfortable with treating curves involving the same variable, x, or y now thanks to you guys. My question now is about finding the region bounded by x=y^2-y and y=x^2+x
I can get the points of intersection, but otherwise, I'm pretty lost. Any help would be greatly appreciated

 

[pka]

 

[NYC]

Thanks pka! I do have one lingering question though: is it possible to do this without approximating the area? I came up with a bunch of ideas, but I wasn't too sure about using the intersection point of .83928676, 1.543689026 Any additonal insight would be great

 

[pka]

Look at the integral that I used!
Like last night, we use the ‘right most’ minus the ‘left most’.
That is one integral not several areas.

Note that I solved \(\displaystyle \L
y = x^2 + x\) as \(\displaystyle \L
x = \frac{{ - 1 + \sqrt {1 + 4y} }}{2}\)

 

[galactus]

Here's a way to go about it. A little 'around the horn'. This is a tricky one. Done some 'head-scratchin':

\(\displaystyle \int\limits_{ - {\textstyle{1 \over 4}}}^{{\textstyle{{14403} \over {17161}}}} {{\textstyle{{\sqrt {4x + 1} + 1} \over 2}} - (x^2 + x)dx} - \int\limits_{{\textstyle{{ - 1} \over 4}}}^0 {{\textstyle{{ - \sqrt {4x + 1} + 1} \over 2}} - (x^2 + x)dx}=.711628165912\)

 

[NYC]

Oof, ok I got it now, thank you very much. The original way i did it was, as you said, a series of areas, but I understand your method; which is much more reliable than mine. Thank you again.