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anonymous
 5 years ago
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
 5 years ago
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you looking for the area?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the formula confuses me, do you do the derivative or the integral first?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So find where they intersect, then integrate the top function  the bottom function between the two x values for their intersection.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they go from x=1 to 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so You're finding the area above y=x^2+1 and below y=2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you just integrated the function y=2 from 1 to 1 what area would you be finding?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0And the region we care about is inside that square, but is smaller than that whole area.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=\int\limits_{1}^{1}[x ^{2}1]dx \] is where I'm at....I follow you so far...just didnt want to delete this equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That is the area of the region _under_ the parabola but above y=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the area would be 4 then, without the parabola

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If 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
 5 years ago
Best ResponseYou've already chosen the best response.0So \[ A_{between} = A_{undertop}  A_{underbottom} \] \[ = \int\limits_{x_0}^{x_1} f_{top}(x)f_{bottom}(x)\ dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, you were confused about how to do the integral?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, I didn't think so, could you show me how to do the integral of the x^2+1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahh I can look it up, thanks for the help though!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{1}2  (x^2+1)dx = \int\limits_{1}^{1}1dx  \int\limits_{1}^{1}x^2dx = x_{1}^{1}  \frac{x^3}{3}_{1}^{1}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I got 4/3...but I have no idea how to do what you just did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well 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
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, this site has bugs...yes I do know that, what is the dx doing at the end though?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's the variable you're integrating with. assume that f and g are functions of x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So you have f = 2, and g = (x^2 + 1) in this case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All 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
 5 years ago
Best ResponseYou've already chosen the best response.0ok with you so far, do you have to graph it out for the integrals?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you 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
 5 years ago
Best ResponseYou've already chosen the best response.0so for the integral of x^2 from 1 to 1 (which is area under the curve) how do you do that?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh. \[\int x^adx = \frac{x^{a+1}}{a+1} + C\]
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