## 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

1. anonymous

Are you looking for the area?

2. anonymous

yup

3. anonymous

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

4. anonymous

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

5. anonymous

they go from x=-1 to 1

6. anonymous

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

7. anonymous

yup

8. anonymous

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

9. 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

10. 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.

11. anonymous

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

12. 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

13. anonymous

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

14. anonymous

the area would be 4 then, without the parabola

15. 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?

16. anonymous

yes it would

17. anonymous

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

18. 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

19. anonymous

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

20. anonymous

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

21. anonymous

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

22. anonymous

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

23. anonymous

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

24. anonymous

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

25. anonymous

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

26. anonymous

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

27. anonymous

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

28. 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.

29. anonymous

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

30. 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.

31. anonymous

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

32. anonymous

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

33. anonymous

ah good thanks