(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I should have to prove that [tex]\int[/tex][tex]^{b}_____________{a}[/tex] x² dx = (1/3)(b³-a³)

3. 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(x_{0})[tex]\Delta[/tex]x_{1}[/tex]+f(x_{1})[tex]\Delta[/tex]x_{2}

+ ... + f(x_{n-1})[tex]\Delta[/tex]x_{n}

U[tex]_{f}[/tex](P) = f(x_{1})[tex]\Delta[/tex]x_{1}+f(x_{2})[tex]\Delta[/tex]x_{2}

+ ... + f(x_{n})[tex]\Delta[/tex]x_{n}

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:

3x_{i-1}² [tex]\leq[/tex]x_{i-1}²+x_{i-1}x_{i}+x_{i}²[tex]\leq[/tex]3x_{i}² "

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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Prove that the integr of x² = (1/3)(b³-a³)

**Physics Forums | Science Articles, Homework Help, Discussion**