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
@amistre64 @TuringTest @saifoo.khan @asnaseer @AccessDenied @Mertsj @JamesJ
If you can help, I would really appreciate it. This is my last question for my e-assignment.
Hey @SMISHRA would you mind helping me out with this one?
Sorry I prefer chemistry over mathematics. @mark1. Personal preference.
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:
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.
So this is where I used the |f''(x)| ≤ K expression
Then, I used the second expression, that inequality with |Em| involved and subbed in K.
a = 0, b = 3.2, and there are.... AHH! How many subintervals are there (ie. what is n????)
I would think n=8 9 values, so 8 intervals between them.
Ok I just realized that K = 5.
You are SUPPOSE TO ABSOLUTE VALUE! It just hit me.
Yeah, I was thinking about that also. -5 <= f"(x) <= 2 |f"(x)| <= 5
Actually, this is what they did. They let n = 4, because the value in between each value is the midpoint. Get it?
Sorry, I meant "in between the 2 values"
Now I am getting 0.1333 if I solve the inequality, but apparently, that is still wrong
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