Question

APPROXIMATE INTEGRATION.
Ok PLEASE CLICK FIRST LINK, then SECOND. Its v.impt that you follow that order:
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36-IMG-004.png
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36/MATA36-IMGa-0002.png

HELP
@amistre64
@TuringTest
@saifoo.khan
@asnaseer
@AccessDenied
@Mertsj
@JamesJ

Answer 1

@amistre64 @TuringTest @saifoo.khan @asnaseer @AccessDenied @Mertsj @JamesJ

Answer 2

If you can help, I would really appreciate it. This is my last question for my e-assignment.

Answer 3

Hey @SMISHRA would you mind helping me out with this one?

Answer 4

Haha

Answer 5

Sorry I prefer chemistry over mathematics. @mark1. Personal preference.

Answer 6

How did you get 0.1707? Like, what were the values of the variables you used?

I'll try to figure it out, if I can. D:

Answer 7

Ok this is what I did. I realized that the highest value that the second derivative takes on is 2. K is the value that is greater than or equal to the absolute value of the second derivative.

Answer 8

So this is where I used the |f''(x)| ≤ K expression

Answer 9

Then, I used the second expression, that inequality with |Em| involved and subbed in K.

Answer 10

a = 0, b = 3.2, and there are.... AHH! How many subintervals are there (ie. what is n????)

Answer 11

I would think n=8
9 values, so 8 intervals between them.

Answer 12

Ok I just realized that K = 5.

Answer 13

You are SUPPOSE TO ABSOLUTE VALUE! It just hit me.

Answer 14

Yeah, I was thinking about that also.
-5 <= f"(x) <= 2
|f"(x)| <= 5

Answer 15

Actually, this is what they did. They let n = 4, because the value in between each value is the midpoint. Get it?

Answer 16

Sorry, I meant "in between the 2 values"

Answer 17

Now I am getting 0.1333 if I solve the inequality, but apparently, that is still wrong