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

- anonymous

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- anonymous

Are you looking for the area?

- anonymous

yup

- anonymous

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

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## More answers

- anonymous

So find where they intersect, then integrate the top function - the bottom function between the two x values for their intersection.

- anonymous

they go from x=-1 to 1

- anonymous

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

- anonymous

yup

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

\[A=\int\limits_{-1}^{1}[x ^{2}-1]dx \] is where I'm at....I follow you so far...just didnt want to delete this equation

- anonymous

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

- anonymous

the area would be 4 then, without the parabola

- anonymous

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?

- anonymous

yes it would

- anonymous

So
\[ A_{between} = A_{undertop} - A_{underbottom} \]
\[ = \int\limits_{x_0}^{x_1} f_{top}(x)-f_{bottom}(x)\ dx\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

\[\int\limits_{-1}^{1}2 - (x^2+1)dx = \int\limits_{-1}^{1}1dx - \int\limits_{-1}^{1}x^2dx = x|_{-1}^{1} - \frac{x^3}{3}|_{-1}^{1}\]

- anonymous

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

- anonymous

Well you know that the integral of a sum is the sum of the integrals right?
\[\int\limits_{a}^{b} (f + g)dx = \int\limits_{a}^{b}f\ dx + \int\limits_{a}^{b}g\ dx\]

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

ah good thanks

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