anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Are you looking for the area?
anonymous
  • anonymous
yup
anonymous
  • anonymous
the formula confuses me, do you do the derivative or the integral first?

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anonymous
  • anonymous
So find where they intersect, then integrate the top function - the bottom function between the two x values for their intersection.
anonymous
  • anonymous
they go from x=-1 to 1
anonymous
  • anonymous
Ok, so You're finding the area above y=x^2+1 and below y=2.
anonymous
  • anonymous
yup
anonymous
  • anonymous
If you just integrated the function y=2 from -1 to 1 what area would you be finding?
anonymous
  • 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
  • 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
  • anonymous
And the region we care about is inside that square, but is smaller than that whole area.
anonymous
  • 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
  • anonymous
That is the area of the region _under_ the parabola but above y=0
anonymous
  • anonymous
the area would be 4 then, without the parabola
anonymous
  • 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
  • anonymous
yes it would
anonymous
  • anonymous
So \[ A_{between} = A_{undertop} - A_{underbottom} \] \[ = \int\limits_{x_0}^{x_1} f_{top}(x)-f_{bottom}(x)\ dx\]
anonymous
  • 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
  • anonymous
Oh, you were confused about how to do the integral?
anonymous
  • anonymous
yeah, I didn't think so, could you show me how to do the integral of the x^2+1?
anonymous
  • anonymous
ahh I can look it up, thanks for the help though!
anonymous
  • 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
  • anonymous
I got 4/3...but I have no idea how to do what you just did
anonymous
  • 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
  • anonymous
sorry, this site has bugs...yes I do know that, what is the dx doing at the end though?
anonymous
  • anonymous
that's the variable you're integrating with. assume that f and g are functions of x
anonymous
  • anonymous
So you have f = 2, and g = -(x^2 + 1) in this case.
anonymous
  • 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
  • anonymous
ok with you so far, do you have to graph it out for the integrals?
anonymous
  • 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
  • anonymous
so for the integral of x^2 from -1 to 1 (which is area under the curve) how do you do that?
anonymous
  • anonymous
Oh. \[\int x^adx = \frac{x^{a+1}}{a+1} + C\]
anonymous
  • anonymous
ah good thanks

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