region bounded by the curves y=x^2+1 and y=2. I know the formula, just don't know how to proceed through it

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region bounded by the curves y=x^2+1 and y=2. I know the formula, just don't know how to proceed through it

Mathematics
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Are you looking for the area?
yup
the formula confuses me, do you do the derivative or the integral first?

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So find where they intersect, then integrate the top function - the bottom function between the two x values for their intersection.
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?
so far I just graphed it out, and they intersect at -1 and 1...looking for the area between -1 and 1, with y=2 on top and y=x^2+1 on the bottom
So to answer my question. integrating y=2 from -1 to 1 would give the area of a square lying on the x axis centered around the line x=0.
And the region we care about is inside that square, but is smaller than that whole area.
\[A=\int\limits_{-1}^{1}[x ^{2}-1]dx \] is where I'm at....I follow you so far...just didnt want to delete this equation
That is the area of the region _under_ the parabola but above y=0
the area would be 4 then, without the parabola
If we take the whole square, and we subtract the area of the region between the parabola and y=0 won't that give us the area we care about?
yes it would
So \[ A_{between} = A_{undertop} - A_{underbottom} \] \[ = \int\limits_{x_0}^{x_1} f_{top}(x)-f_{bottom}(x)\ dx\]
right, I knew that equation, it was how to go through it, but you also answered that question...it's the integral that I had forgotten how to do
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!
\[\int\limits_{-1}^{1}2 - (x^2+1)dx = \int\limits_{-1}^{1}1dx - \int\limits_{-1}^{1}x^2dx = x|_{-1}^{1} - \frac{x^3}{3}|_{-1}^{1}\]
I got 4/3...but I have no idea how to do what you just did
Well you know that the integral of a sum is the sum of the integrals right? \[\int\limits_{a}^{b} (f + g)dx = \int\limits_{a}^{b}f\ dx + \int\limits_{a}^{b}g\ dx\]
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.
All I did was take the -1 from g, and add it to the 2 in f, then break it up into two separate integrals.
ok with you so far, do you have to graph it out for the integrals?
you can simply find where they intersect and use those for your limits, then subtract the integral of the bottom function from the integral of the top function.
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

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