Prove that the integr of x² = (1/3)(b³-a³)

[maupassant]

Homework Statement

I should have to prove that $$\int$$$$^{b}_____________{a}$$ 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$$_{f}$$(P)$$\leq$$ I $$\leq$$ U$$_{f}$$(P) .

So the first thing to do is to find the lower and upper sum which are:
L$$_{f}$$(P) = f(x₀)$$\Delta$$x₁[/tex]+f(x₁)$$\Delta$$x₂
+ ... + f(x_n-1)$$\Delta$$x_n

U$$_{f}$$(P) = f(x₁)$$\Delta$$x₁+f(x₂)$$\Delta$$x₂
+ ... + f(x_n)$$\Delta$$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-1² $$\leq$$x_i-1²+x_i-1x_i+x_i²$$\leq$$3x_i² "

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!

 

[maupassant]

Hmm should 've used subscripts instead of superscripts... Apologies!

 

[dynamicsolo]

That middle term looks very much like one of the factors in a difference of two cubes:

$$a^3 - b^3 = (a - b) \cdot (a^2 + ab + b^2)$$

 

[maupassant]

Thank you for the quick reply but the thing that bothers me is how one reaches
inequality 3x_i-1<=x_i²+x_i-1x_i<=3x_i

And why did the author multiply the first and the third factor by 3? I cannot give any reason to do so unless perhaps you work from the result (=(1/3)(b³-a³)) backwards? So, how would the author know how to proceed further from: U_f(P) and L_f(P) ?


Thank you.

 

[dynamicsolo]

I suspect a certain measure of hokum in this proof, in the sense that the party arguing knew what they wanted to prove before they proved it.

I haven't sorted all of this out yet, but the inequality

$$3x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1}x_{i} +x_{i}^2 \leq 3x_{i}^2$$

comes from the idea that there are three terms in the middle, two of which are plainly larger than $$x_{i-1}^2$$ and two of which are smaller than $$x_{i}^2$$. If you now divide through by 3 and multiply all terms by $$(x_{i} - x_{i-i})$$, which is the $$\Delta x$$, you have the factors for the terms of the Riemann sum for the lower and upper sum on each end and the factors for $$\frac{1}{3}(x_{i}^3 - x_{i-1}^3)$$ in the middle. So you have the term you're after bracketed in the inequality, thus the sum over i on this inequality will bracket $$\frac{1}{3}(b^3 - a^3)$$, since all the middle terms in that sum will cancel out.

Not having your book in front of me, I don't know the full context of this. It's a clever argument, but also one where it helps to know what you're intending to come up with. (You see that at times in proofs in textbooks.)