Math: Limit Problem, one puzzling question.

[Soontir C'boath]

Ok, I'm studying for an exam coming up soon but I am looking at a question right now that doesn't follow what I read.

The problem reads:

Express the area under the curve y = x^5 from 0 to 2 as a limit.

That's what I did and I figure it to be the limit as n->infinity (2/n) Sigma i^5

Except instead of (2/n) the book's answer has it with an exponet of six making it (64/n^6).

Isn't it just (b-a)/n? Does it have to do with perhaps antideriving(which i have no idea, just throwing a guess) x^5?

[Count Dooku]

I'm in my first semester of Differential Calculus, so I don't think I'd be much help. Your book SHOULD have a few Theorum's for this. You might have to use more than one to to solve the problem. Again, sorry for the lack of help.

[Beowulf]

Area under a curve = integral, or anti-derivative. That's nearly the definition of integral.

[Soontir C'boath]

I'm in my first semester of Differential Calculus, so I don't think I'd be much help. Your book SHOULD have a few Theorum's for this. You might have to use more than one to to solve the problem. Again, sorry for the lack of help.

That's in two chapters after this one.

Area under a curve = integral, or anti-derivative. That's nearly the definition of integral.

Well, yea I know that, but why does it apply an exponet of 6 to what is perhaps considered a constant that shouldn't have anything to do with it it.

[Sriad]

Indeed. Writing it as a limit as n->infinity is making a lot more work for yourself than the question is asking for, you just need a limit as x->2 unless you left some reeeally mean stipulation out rephrasing the question.

[Soontir C'boath]

Indeed. Writing it as a limit as n->infinity is making a lot more work for yourself than the question is asking for, you just need a limit as x->2 unless you left some reeeally mean stipulation out rephrasing the question.

That's how it was, though the reading section does said not to evaluate it in an example it had in which both are similar.

[Durandal]

Ok, I'm studying for an exam coming up soon but I am looking at a question right now that doesn't follow what I read.

The problem reads:

Express the area under the curve y = x^5 from 0 to 2 as a limit.

That's what I did and I figure it to be the limit as n->infinity (2/n) Sigma i^5

The proper anti-derivative would be x^6 / 6 from 0 to 2.

Except instead of (2/n) the book's answer has it with an exponet of six making it (64/n^6).

Isn't it just (b-a)/n? Does it have to do with perhaps antideriving(which i have no idea, just throwing a guess) x^5?

Integration is the limit of a Reimann sum as n->infinity. A Reimann sum is the summation from x_i to x_f of f(x_i) * delta-x. In other words, the result of the function's value at x_i multiplied by your step size, which gives you the area of a rectangle of height f(x_i) and width delta-x.

To express the area under a curve, you have to let the number of these rectangles go to infinity to get infinite precision. So your step size is going to get as close to zero as possible. So the Reimann sum will look something like this

limit(n->infinity) Sum(i=0, i=n) f(x_i) * delta-x
limit(n->infinity) Sum(i=0, i=n) (x_i)^5 * delta-x
limit(n->infinity) Sum(i=0, i=2) (x_i)^5 * ((2 - 0) / n)
limit(n->infinity) Sum(i=0, i=2) (2 * (x_i)^5) / n

That should get you started. Now just take that limit as n->infinity, and you should have your answer.

[Kuroneko]

Ok, I'm studying for an exam coming up soon but I am looking at a question right now that doesn't follow what I read. The problem reads:

Express the area under the curve y = x^5 from 0 to 2 as a limit.

That's what I did and I figure it to be the limit as n->infinity (2/n) Sigma i^5. Except instead of (2/n) the book's answer has it with an exponet of six making it (64/n^6).

Partition the interval [0,2] into n subintervals, say, I_k = {[a_0,a_1)...,[a_k,a_{k+1}),...,[a_{n-1},a_n]}, where a_k = 2k/n, which has all of them length 2/n. The area under the curve can be approximated as the sum of all rectangles determined by the subintervals with height x_k^5, where x_k is in I_k. Since the function is continuous on [0,2], absolutely any choice of x_k will do. The relevant limit is then:
lim_{n→∞} Sum_{0≤k<n}[ (2/n) x_k^5 ]
A particularly nice choice is x_k = a_k = 2k/n, giving
lim_{n→∞} Sum_{1≤k<n}[ (2/n) (2k/n)^5 ].
If you look at the summation itself, n is constant, so you can pull it out. That's where the (64/n^6) comes from.

[Durandal]

I shall, as always, defer to Kuroneko.

[Soontir C'boath]

I definitely see it now. Since (x_i)^5 is basically in the summation when replacing f(x), most of the points are equal to (2i/n) which is conveniantly not at a point such as (2n/n). (2i/n) can then easily replace (x_i) and then the exponetial factor of five takes care of it.

Does it change at all if the interval itself goes to infinity?

It still have subintervals that will equal to (2i/n) and the function can still be replaced by it at those points although at an infinite amount, right?

Thank you, Kuroneko, Durandel, and the rest of you gets a cookie.

[Braedley]

To express the area under a curve, you have to let the number of these rectangles go to infinity to get infinite precision. So your step size is going to get as close to zero as possible. So the Reimann sum will look something like this

limit(n->infinity) Sum(i=0, i=n) f(x_i) * delta-x
limit(n->infinity) Sum(i=0, i=n) (x_i)^5 * delta-x
limit(n->infinity) Sum(i=0, i=2) (x_i)^5 * ((2 - 0) / n)
limit(n->infinity) Sum(i=0, i=2) (2 * (x_i)^5) / n

That should get you started. Now just take that limit as n->infinity, and you should have your answer.

Yeah, that doesn't look quite right. From what I remember from 2+ years ago:

where
and

[Durandal]

Yeah I went from 0 to 2, rather than 1 to n, like an idiot.

[Kuroneko]

I definitely see it now. Since (x_i)^5 is basically in the summation when replacing f(x), most of the points are equal to (2i/n) which is conveniantly not at a point such as (2n/n). (2i/n) can then easily replace (x_i) and then the exponetial factor of five takes care of it.

There's nothing problematic about 2n/n. In fact, if the partition was chosen a bit differently, with x_k = 2(k+1)/n instead, we'd have lim_{n→∞} Sum_{0≤k<n}[ (2/n) [2(k+1)/n]^5 ]. Note that the last term of that summation is (2/n)(2n/n)^5.

Does it change at all if the interval itself goes to infinity?

For [a,∞), compute it at a finite interval [a,b] and take limit as b→∞. The Riemann integral generalizes very poorly to infinite intervals.

It still have subintervals that will equal to (2i/n) and the function can still be replaced by it at those points although at an infinite amount, right?

Think in terms of partitions. How would partition an infinite interval, say [0,∞), into finitely many pieces without having at least one infinite subinterval? Normally, the assumption of continuity allows you to pick any x_k inside a subinterval, since as the lengths of the subintervals tend to zero in the limit, different choices will tend to the same function value anyway. You don't have that if you allow an infinite subinterval. If this was a Riemann-Stieltjes integral, you could do something more creative with partitions, but even that is dangerous. The finite [0,b] with limit b→∞ avoids the problem cleanly.