Prove the integral of f over [0,1] is 1. f=(n+1)x^n when 0<=x<1; f=1, when x=1

[mathdunce]

Homework Statement

Homework Equations

The Attempt at a Solution

It is very easy to see intuitively that \int_0^1 f=1, because the endpoint x=1 has no impact on the integral. If we see the integral as the area under the curve, then the area under f(1) is zero. However, we are required to prove this rigorously. It is really painful. I tried the definition of Riemann sum and the Cauchy criterion for integrability, but failed. I thought this exercise would be extremely easy, but failed to solve it after great efforts. Could anyone please give me a hint or a solution? Thank you!

 

[Coto]

I believe it's a well known theorem that a function is Riemann integrable iff it's bounded and continuous except on a set of measure zero. Since the end point is a set of measure zero, there's nothing much left to show. I don't believe the question is asking for a proof of this theorem, only that you apply it.

 

[mathdunce]

I believe it's a well known theorem that a function is Riemann integrable iff it's bounded and continuous except on a set of measure zero. Since the end point is a set of measure zero, there's nothing much left to show. I don't believe the question is asking for a proof of this theorem, only that you apply it.

Coto, Thank you! I agree.

 

[brydustin]

Coto is right about the point being "measure zero", but don't worry about that. The question at this point is, "What class is this for?". Depending on the context the professor will expect a different level of rigour. For example, judging my your evaluation (which is wrong), I would assume that this is not an analysis course, but rather simply a calculus one course. Is this correct? But if its an intro to real analysis course, you really have to "feel" out the prof., ask yourself what they want and want you to show. For example, how important is it for the professor to know that you know that the point at the end has no affect on the integration? For example, you ACTUALLY can't make such a statement without backing it up. Not to worry I remember this proof:

http://ocw.mit.edu/courses/mathematics/18-101-analysis-ii-fall-2005/lecture-notes/lecture8.pdf

Use Theorem 3.7, definition 3.9, and Theorem 3.10 to first prove the conditions of integrability of this function.
Now as for the actual integration:

∫_0^1〖f_n dx= ∫_0^1▒〖(n+1) x^n dx=1/(n+1) x^n(n+1) ⋮■(1@0) (evaluated over 0 to 1) 〗〗=
because you can^' t evaluate for x=1 (this would make an ill result),
take the following to do the evalutation:
lim┬(x→1^+ )⁡〖1/(n+1) x^n(n+1) - 〗 lim┬(x→0)⁡〖1/(n+1) x^n(n+1) 〗=1/(n+1)-0
lim┬(x→1^- )⁡〖1/(n+1) x^n(n+1) - 〗 lim┬(x→0)⁡〖1/(n+1) x^n(n+1) =1/(n+1)-0 〗
So remember that the limit of a function may not necessarily be the value of a function at that point. Now compare the left-hand and right-hand limits and notice that they are equal, this means that the integral doesn’t care about that point. Also in your attempt you got that the integral equalled 1, I’m assuming that you evaluated for n and not x (for zero and one in the integral). The value of n is merely a “family” of functions, in other words there are a family of solutions.
∴∫_0^1▒〖f_n (x)dx= 1/(x+1) ∎〗

Homework Statement

Homework Equations

The Attempt at a Solution

It is very easy to see intuitively that \int_0^1 f=1, because the endpoint x=1 has no impact on the integral. If we see the integral as the area under the curve, then the area under f(1) is zero. However, we are required to prove this rigorously. It is really painful. I tried the definition of Riemann sum and the Cauchy criterion for integrability, but failed. I thought this exercise would be extremely easy, but failed to solve it after great efforts. Could anyone please give me a hint or a solution? Thank you!

 

[brydustin]

You should really realize that the function is not evaluted as 1... rather 1/(n+1)

 

[mathdunce]

Coto is right about the point being "measure zero", but don't worry about that. The question at this point is, "What class is this for?". Depending on the context the professor will expect a different level of rigour. For example, judging my your evaluation (which is wrong), I would assume that this is not an analysis course, but rather simply a calculus one course. Is this correct? But if its an intro to real analysis course, you really have to "feel" out the prof., ask yourself what they want and want you to show. For example, how important is it for the professor to know that you know that the point at the end has no affect on the integration? For example, you ACTUALLY can't make such a statement without backing it up. Not to worry I remember this proof:

http://ocw.mit.edu/courses/mathematics/18-101-analysis-ii-fall-2005/lecture-notes/lecture8.pdf

Use Theorem 3.7, definition 3.9, and Theorem 3.10 to first prove the conditions of integrability of this function.
Now as for the actual integration:

∫_0^1〖f_n dx= ∫_0^1▒〖(n+1) x^n dx=1/(n+1) x^n(n+1) ⋮■(1@0) (evaluated over 0 to 1) 〗〗=
because you can^' t evaluate for x=1 (this would make an ill result),
take the following to do the evalutation:
lim┬(x→1^+ )⁡〖1/(n+1) x^n(n+1) - 〗 lim┬(x→0)⁡〖1/(n+1) x^n(n+1) 〗=1/(n+1)-0
lim┬(x→1^- )⁡〖1/(n+1) x^n(n+1) - 〗 lim┬(x→0)⁡〖1/(n+1) x^n(n+1) =1/(n+1)-0 〗
So remember that the limit of a function may not necessarily be the value of a function at that point. Now compare the left-hand and right-hand limits and notice that they are equal, this means that the integral doesn’t care about that point. Also in your attempt you got that the integral equalled 1, I’m assuming that you evaluated for n and not x (for zero and one in the integral). The value of n is merely a “family” of functions, in other words there are a family of solutions.
∴∫_0^1▒〖f_n (x)dx= 1/(x+1) ∎〗

Hello, brydustin. Thank you for your detailed reply. I have finished the exercise by using the following theorem. If g is continuous and f is different from g at a finite number of points then f is integrable and the integral is the same as ∫g. Of course, here we are talking about the integration over a certain closed interval. Again, thanks a lot.