## anonymous one year ago NUmber 8

1. anonymous

Hi!! What is the Number 8 for?

2. misty1212

$\infty$

3. anonymous

$if \left(\begin{matrix}8-x^2 for -2\le x \le2 \\ x^2 else where,\end{matrix}\right) then \int\limits_{-1}^{3}f(x) dx is a number \between$

4. anonymous

is a number between?* would you like to know the choices?

5. anonymous

how would i do this?

6. anonymous

ignore number 8

7. xapproachesinfinity

hmm i see peiewize function and you the area under its curve

8. anonymous

it is a peicewise

9. xapproachesinfinity

what 8 do yu want us to ignore? 8-x^2? this should be part of the funtion

10. anonymous

i am talking about the title of the question. that is what you should ignore

11. xapproachesinfinity

eh soka!

12. misty1212

$f(x) = \left\{\begin{array}{rcc} 8-x^2 & \text{if} & -2\leq x\leq 2\\ x^2& \text{otherwise} & \end{array} \right.$ just seeing if i could do it, ignore me

13. xapproachesinfinity

alright you just do some linearity here see where you should integrate 8-x^2 and x^2

14. xapproachesinfinity

as you can see from -1<x<2 you can integrate 8-x^2

15. xapproachesinfinity

the rest 2 to 3 you integrate x^2

16. xapproachesinfinity

do you get it, or should i write it down?

17. anonymous

i know the answer is 8

18. anonymous

19. xapproachesinfinity

$\int_{-1}^{3}f(x)dx=\int_{-1}^{2}f(x)dx+\int_{2}^{3}f(x)dx$ $\int_{-1}^{2}(8-x^2)dx+\int_{2}^{3}x^2dx$

20. anonymous

The choices were a) 0 and 8 b) 8 and 16 c) 16 and 24 d) 24 and 32 e) 32 and 40

21. xapproachesinfinity

hmm i see so they separated it you just do those two integral separately

22. xapproachesinfinity

$\int_{-1}^{2}(8-x^2)dx=?$

23. anonymous

43/3

24. xapproachesinfinity

really how ?

25. anonymous

the integral is 8x-(1/3)x^3

26. xapproachesinfinity

i found 62/3

27. anonymous

i tried it again, it gave me 21

28. anonymous

29. xapproachesinfinity

yes 21 is good!

30. xapproachesinfinity

but then it is not in the choices

31. anonymous

I believe the answer is B

32. anonymous

Earlier,I said the solution to the intergral was 8 and i was wrong

33. xapproachesinfinity

something is wrong in what you wrote? that way must work

34. xapproachesinfinity

can you post a snap shot of the problem please

35. xapproachesinfinity

is the function exately like missy wrote it

36. xapproachesinfinity

hey yo there?

37. xapproachesinfinity

???

38. xapproachesinfinity

i'm waiting...

39. anonymous

sorry, i will do that now

40. anonymous

41. xapproachesinfinity

now i got what is those btw mean we need to add those two integrals and see the solution lies btw what two points

42. xapproachesinfinity

i found 21 +19/3 =82/3 which lies 24<82/3<32

43. xapproachesinfinity

go with what we found in the first time and add the other integral

44. xapproachesinfinity

see if you got same as me

45. xapproachesinfinity

just miss understood the problem

46. anonymous

i got 21-(19/3)

47. xapproachesinfinity

hmm why -19/3?

48. anonymous

never mind

49. anonymous

we are good.

50. anonymous

so D?

51. xapproachesinfinity

$\int_{2}^{3}x^2ds=\frac{x^3}{3}=19/3$

52. xapproachesinfinity

yes D

53. anonymous

Instead of getting the sum of both integrals, I decided to do the difference.

54. xapproachesinfinity

yeah i see that :)

55. xapproachesinfinity

alright i think we solved that :)

56. anonymous

Arigato!

57. xapproachesinfinity

No problem :) yakoso