Real Analysis

[Brainwave]

Pls help on this problem.Let F: [0,a] ----> Real number be defined byX ----> x^2, Using partition unit sub-intervalsP = { [0,a/n],[a/n,2a/n],...,[(2a(n-1))/n , a]}.Obtain integral F(x) from 0 to a. Pls F is Riemann integrable. I need your help.

 

[JeffM]

Pls help on this problem.Let F: [0,a] ----> Real number be defined byX ----> x^2, Using partition unit sub-intervalsP = { [0,a/n],[a/n,2a/n],...,[(2a(n-1))/n , a]}.Obtain integral F(x) from 0 to a. Pls F is Riemann integrable. I need your help.

Brainwave

You have asked questions before and know how we work. What have you done (if anything)? If you were not even able to start, do you know what a Riemann sum is? Do you know how to express such a sum?

 

[Brainwave]

Yes, It is the partition sub-intervals that gives problem. I don't really knw how to go about the sub-intervals.

 

[JeffM]

Yes, It is the partition sub-intervals that gives problem. I don't really knw how to go about the sub-intervals.

OK. Let's take this in very small steps. For our first approximation, let n = 2, and let's use the right side of rectangles. (We could use trapezoids, or mid-points of rectangles or left sides of rectangles: it will make no difference in the limit as n approaches infinity.)

The function to be integrated is f(x) = x^2.

Our first rectangle has a base from 0 to (a / 2). Along the right side, its height is (a / 2)^2 so the area of that rectangle is

\(\displaystyle A_1 = \left(\dfrac{a}{2} - 0\right) * \left(\dfrac{a}{2}\right)^2 = \dfrac{a}{2} * \dfrac{a^2}{4} = \dfrac{a^3}{8}.\)

If you do not understand that, I suggest you draw a sketch. Feel free to ask questions if you still do not get it.

Now the base of our second rectangle runs from (a / 2) to a, and the height is a^2, right?

\(\displaystyle A_2 = \left(a - \dfrac{a}{2}\right) * a^2 = \dfrac{a}{2} * a^2 = \dfrac{a^3}{2}.\)

So the approximate area under the curve is

\(\displaystyle A_1 + A_2 = \dfrac{a^3}{8} + \dfrac{a^3}{2} = \dfrac{a^3}{8} + \dfrac{4a^3}{8} = \dfrac{5a^3}{8} > \dfrac{a^3}{2}.\)

So that is not a super approximation, but it was very crude. Let's try three subintervals, which gives three rectangles.

\(\displaystyle A_1 = \left(\dfrac{a}{3} - 0\right) * \left(\dfrac{a}{3}\right)^2 = \dfrac{a}{3} * \dfrac{a^2}{9} = \dfrac{a^3}{27}.\)

\(\displaystyle A_2 = \left(\dfrac{2a}{3} - \dfrac{a}{3}\right) * \left(\dfrac{2a}{3}\right)^2 = \dfrac{a}{3} * \dfrac{4a^2}{9} = \dfrac{4a^3}{27}.\)

\(\displaystyle A_3 = \left(a - \dfrac{2a}{3}\right) * a^2 = \dfrac{a}{3} * a^2 = \dfrac{a^3}{3}.\)

So the approximate area under the curve is

\(\displaystyle A_1 + A_2 + A_3 = = \dfrac{a^3}{27} + \dfrac{4a^3}{27}+ \dfrac{a^3}{3} = \dfrac{5a^3}{27} + \dfrac{9a^3}{27} = \dfrac{14a^3}{27} > \dfrac{a^3}{2}.\)

That is actually a pretty decent approximation. Now if you had n rectangles and added up their areas, how would you express that in algebraic form?

What is the limit of that if you let n approach infinity?

 

[pka]

Yes, It is the partition sub-intervals that gives problem. I don't really knw how to go about the sub-intervals.

This is a left-hand sum: \(\displaystyle \displaystyle\sum\limits_{k = 0}^{n - 1} {{{\left( {\frac{{ak}}{n}} \right)}^2}\left( {\frac{a}{n}} \right)} = \left( {\frac{{{a^3}}}{{{n^3}}}} \right)\sum\limits_{k = 0}^{n - 1} {{k^2}} \)

This is a right-hand sum: \(\displaystyle \displaystyle\sum\limits_{k = 1}^{n } {{{\left( {\frac{{ak}}{n}} \right)}^2}\left( {\frac{a}{n}} \right)} = \left( {\frac{{{a^3}}}{{{n^3}}}} \right)\sum\limits_{k =1}^{n} {{k^2}} \)

This is an edit and post script in response to reply #4.
Given an increasing function on \(\displaystyle [0,a]\) such as \(\displaystyle x^2\) then any left-hand sum is a lower sum and any right-hand sum is an upper sum.

Here is a special sum: \(\displaystyle \sum\limits_{k = 1}^n {{k^2}} = \frac{{n(n + 1)(2n + 1)}}{6}\).

With that, the two sums above, and the squeeze theorem show that \(\displaystyle \displaystyle{\lim _{n \to \infty }}Sum = \frac{{{a^3}}}{3}\)

 

[JeffM]

This is an edit and post script in response to reply #4.
Given an increasing function on \(\displaystyle [0,a]\) such as \(\displaystyle x^2\) then any left-hand sum is a lower sum and any right-hand sum is an upper sum.

Here is a special sum: \(\displaystyle \sum\limits_{k = 1}^n {{k^2}} = \frac{{n(n + 1)(2n + 1)}}{6}\).

With that, the two sums above, and the squeeze theorem show that \(\displaystyle \displaystyle{\lim _{n \to \infty }}Sum = \frac{{{a^3}}}{3}\)

Yes, yours is undoubtedly a much more elegant and rigorous explanation than mine, which was nothing more than an unsupported and conclusory statement that different approaches will somehow magically end up in the same place in the limit. I should have just skipped the whole issue because it is unlikely to be helpful as yet for someone who, apparently, cannot set up any kind of sum. But thanks for the correction. It taught me how to respond better in the future.