FInd the area of region enclosed by the given curves y=x+5 y^2=x y=2 y=-1 what I did was sliced it to parts. from o to -1 and 0 to 2 did I do it right? top half delta y and bottom delta x

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FInd the area of region enclosed by the given curves y=x+5 y^2=x y=2 y=-1 what I did was sliced it to parts. from o to -1 and 0 to 2 did I do it right? top half delta y and bottom delta x

Mathematics
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hello are you here? I would start off with a drawing first
\[\int\limits_{-1}^{2} (\text{ \right }-\text{ \left } )dy\]
yes I understand that and I wanted to split it in half at y=0 and take areas separately would that work? @freckles

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yes but why do that?
seems much easier just to do one integral \[\int\limits_{-1}^2 (y^2-(y-5)) dy\]
I guess I am just over thinking it. because today my teacher was talking about splitting integrals too for some equations
well you have a x=y^2 is right of x=y-5 so x=y^2 is greater than x=y-5
|dw:1442380700153:dw|
it is clear from the picture between y=-1 and y=2 we have x=y^2 is greater than x=y-5 you only need one integral
if the functions had crossed passed and therefore you had some kind of switching of which was greater than you may need more than one integral
for example if you had: |dw:1442380823519:dw| you would need two integrals here
|dw:1442380886729:dw| \[\int\limits_{-1}^c (g(y)-f(y))dy+\int\limits_c^2 (f(y)-g(y))\]
ahhhhhhhh I see and yeeah that looks like two different areas for the second example lol thank you!!
oops forgot to write a dy at the end of that one integral
\[\int\limits\limits_{-1}^c (g(y)-f(y))dy+\int\limits\limits_c^2 (f(y)-g(y)) \color{red}{ dy}\]

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