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Are you looking for the area?

yup

the formula confuses me, do you do the derivative or the integral first?

they go from x=-1 to 1

Ok, so
You're finding the area above y=x^2+1 and below y=2.

yup

If you just integrated the function y=2 from -1 to 1 what area would you be finding?

And the region we care about is inside that square, but is smaller than that whole area.

That is the area of the region _under_ the parabola but above y=0

the area would be 4 then, without the parabola

yes it would

Oh, you were confused about how to do the integral?

yeah, I didn't think so, could you show me how to do the integral of the x^2+1?

ahh I can look it up, thanks for the help though!

I got 4/3...but I have no idea how to do what you just did

sorry, this site has bugs...yes I do know that, what is the dx doing at the end though?

that's the variable you're integrating with. assume that f and g are functions of x

So you have f = 2, and g = -(x^2 + 1) in this case.

ok with you so far, do you have to graph it out for the integrals?

so for the integral of x^2 from -1 to 1 (which is area under the curve) how do you do that?

Oh.
\[\int x^adx = \frac{x^{a+1}}{a+1} + C\]

ah good thanks