- #1
maupassant
- 10
- 0
Homework Statement
I should have to prove that [tex]\int[/tex][tex]^{b}_____________{a}[/tex] x² dx = (1/3)(b³-a³)
The Attempt at a Solution
I know that f is continuous on [a,b] and that there exists a unique number I so that
L[tex]_{f}[/tex](P)[tex]\leq[/tex] I [tex]\leq[/tex] U[tex]_{f}[/tex](P) .
So the first thing to do is to find the lower and upper sum which are:
L[tex]_{f}[/tex](P) = f(x0)[tex]\Delta[/tex]x1[/tex]+f(x1)[tex]\Delta[/tex]x2
+ ... + f(xn-1)[tex]\Delta[/tex]xn
U[tex]_{f}[/tex](P) = f(x1)[tex]\Delta[/tex]x1+f(x2)[tex]\Delta[/tex]x2
+ ... + f(xn)[tex]\Delta[/tex]xn
Now that we 've found the Upper and Lower sums, we must try to find the number I.
My textbook says the following:
" For each index i, 1<=i<=n we have:
3xi-1² [tex]\leq[/tex]xi-1²+xi-1xi+xi²[tex]\leq[/tex]3xi² "
Now where does the author get the middle term from? Why can't one just the take the mean like this (, or does the middle term means something else?) : f((b+a)/2) ?
But apparently this doesn't lead to the end result of the middle term , i.e. (1/3)(b³-a³).
How does one reach this end result from the Upper and Lower sums?
Thank for your help!
Last edited: