#1
Suppose P = {x0,x1,x2, ··· ,xn} is a partition of f (x). Consider the partition Q = {x0,x1, ··· ,xk−1, t,xk, ··· ,xn}.
Notice that Q is simply the partition P but with just one more point added. It should make sense that Lf (P) <= Lf (Q). Let’s prove this formally.

(a) Let m' = min{ f (x) : xk−1 <= x <= t} and m'' = min{ f (x) : t <= x <= xk}. Prove that Lf (P) <= Lf (Q) if and
only if mk(xk −xk−1) <= m' (t −xk−1)+m'' (xk −t).
(b) Now prove that mk(xk −xk−1) <= m'(t −xk−1)+m''(xk −t), thereby completing the proof that Lf (P) <= Lf (Q). (What can you say about mk compared to m', and mk compared to m''?)
(c) Let P and Q be the partitions as given above. What can you say about the upper sums? Deduce a similar
statement and give a proof.

Notes: m' = first derivative of m. m'' = second derivative of m. xk = x sub k.

Help, its due in 3 days
#2
wow, what level or grade of math is that??
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#3
im asian and i dont even know how to do that
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#4
write an essay on how you'll never use that in real life. ever. if you do it really well, your teacher should give you credit for being amusing. I know I would give you credit for that.
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#8
Man my english are pretty shitty when it comes to scientific horology,but if you could translate it in greek I bet I oculd help you...lol
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