Area of curves.
Help! The first one is a pain.
@zepdrix I've gotten answers all over the place trying to calculate it. The other two I got really fast :|
I also got 1616 I'm doing: 8(36)+8(36)+8(32)+8(28)+8(24)+18(16)
So for L6 we have 8f(0) + 8f(8) + 8f(16) + 8f(24) + 8f(32) + 8f(40) Those are the 6 rectangles ya? :o Let's factor the 8 out, \(\rm 8\left[f(0)+f(8)+f(16)+f(24)+f(32)+f(40)\right]\) So we'll need to read this graph carefully, hmm
\[\rm 8\left[36+f(8)+f(16)+f(24)+f(32)+f(40)\right]\]For the second height... hmm 35 I think?
\[\rm 8\left[36+35+f(16)+f(24)+f(32)+f(40)\right]\]
how did you get 35? o.o so specifically? oo
so f(16) would be 33?
Ya that sounds about right :) It's just above the 32 mark.
then 29,24,16?
mmm ya
and the answer would come out to 1384?
Yes, that's what I'm getting also. Have more than one guess to burn? I hope? :) lol
Yeah I do x) But we won't be needing them! :) thank youu
yay \c:/
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