Question

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

Answer 1

Are you looking for the area?

Answer 2

yup

Answer 3

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

Answer 4

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

Answer 5

they go from x=-1 to 1

Answer 6

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

Answer 7

yup

Answer 8

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

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

the area would be 4 then, without the parabola

Answer 15

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?

Answer 16

yes it would

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

Well you know that the integral of a sum is the sum of the integrals right?

\[\int\limits_{a}^{b} (f + g)dx = \int\limits_{a}^{b}f\ dx + \int\limits_{a}^{b}g\ dx\]

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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.

Answer 29

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

Answer 30

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.

Answer 31

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

Answer 32

Oh.

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

Answer 33

ah good thanks