Proving Positive Integral using Sums

In summary, the integral from 0 to x of (sin t)/(t + 1) dt is a function of x, f(x), with f(0) = 0. By considering the areas of each arch of the function, it can be shown that the integral is positive. This can be seen by noting that the integral from 0 to pi is positive, and the integral from pi to 2pi is negative but less in absolute value than the positive part. Similarly, the integral from 2pi to 3pi is positive, and the integral from 3pi to 4pi is negative but less in absolute value than the positive part. This pattern continues, showing that the integral is positive and increasing,
  • #1
SpringPhysics
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0

Homework Statement


Prove that the integral from 0 to x of (sin t)/(t + 1) dt > 0.
(Sorry I don't know how to use the integral latex.)

Homework Equations


We have only learned about lower and upper sums, and how the integral is equal to the supremum of lower sums and the infimum of upper sums when sup = inf

The Attempt at a Solution


I took a partition from 0 to x with equal intervals, so that each interval has length x/n for n intervals. I found the lower sum and upper sum, with a difference of x/n * (sin x)/(x + 1). I have no idea what to do afterwards. Am I to eliminate the possibility of the integral being 0 and negative?

EDIT: I "found" the lower and upper sums incorrectly - in fact, they are not possible to determine.
 
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  • #2
Your integral is a function of x, f(x), with f(0) = 0. If you can show that f'(x) > 0 then you can conclude that f is increasing, hence its integral will be positive.

Here's the LaTeX for your integral.
[tex]\int_0^x \frac{sin t}{t + 1}dt[/tex]
 
  • #3
Mark44 said:
Your integral is a function of x, f(x), with f(0) = 0. If you can show that f'(x) > 0 then you can conclude that f is increasing, hence its integral will be positive.

Wouldn't that exclude the decreasing portion of the function for x > 0, since sine is periodic? Moreover, it wouldn't show that the difference between the area of the first and second arcs is positive.
 
  • #4
The function I'm talking about is not (sin t)/(t + 1), which does oscillate as you describe. The one I'm talking about is the integral. The area under the first arch is positive, and is more positive than the negative "area" above the second arch, so overall for the two arches, the integral is positive. It's sort of like adding 5 - 2.5 + 1.25 - .625 + - ... and so on. Each partial sum is positive, despite half the terms being negative.

I believe the approach you need to take is as I described in my previous post.
 
  • #5
Mark44 said:
The function I'm talking about is not (sin t)/(t + 1), which does oscillate as you describe. The one I'm talking about is the integral. The area under the first arch is positive, and is more positive than the negative "area" above the second arch, so overall for the two arches, the integral is positive. It's sort of like adding 5 - 2.5 + 1.25 - .625 + - ... and so on. Each partial sum is positive, despite half the terms being negative.

I believe the approach you need to take is as I described in my previous post.

Ohhhh. I will try that then, thanks.

I got
lim as h -> 0 of [tex]\frac{f(x + h) - f(x)}{h}[/tex] =
lim as h -> 0 of [tex]\frac{\int_x^(x+h) \frac{sin t}{t + 1}dt}{h}[/tex]

I'm stuck here, as I tried to equate the limit to a constant c greater than zero, then using the delta-epsilon definition of the derivative. Then I get a delta dependent on the value of x, which means that the function is not uniformly continuous (but it must be because it is continuous in a closed interval).
 
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  • #6
The integral of sin(t)/(1+t) is positive from 0 to pi. No doubt about that, right? Now the integral from pi to 2pi is negative. But it will be less in absolute value than the positive part of the integral from 0 to pi. Since 1/(1+t) is less on the interval [pi,2pi] than it is on [0,pi]. Similarly the integral on [2pi,3pi] is positive, integral on [3pi,4pi] is negative but less in absolute value than the former. Etc etc. That's what Mark44 was talking about. It's not really a derivative problem.
 
  • #7
Dick said:
The integral of sin(t)/(1+t) is positive from 0 to pi. No doubt about that, right? Now the integral from pi to 2pi is negative. But it will be less in absolute value than the positive part of the integral from 0 to pi. Since 1/(1+t) is less on the interval [pi,2pi] than it is on [0,pi]. Similarly the integral on [2pi,3pi] is positive, integral on [3pi,4pi] is negative but less in absolute value than the former. Etc etc. That's what Mark44 was talking about. It's not really a derivative problem.

But that's not really a proof right? That's just an intuitive solution.
Also, we have not learned how to differentiate an integral.
 
  • #8
It's not just an 'intuitive solution'. It's an informal description of a legitimate proof. How formal you want to make it depends on the requirements of your class. I think if you understand that and describe it clearly, that's all you need.
 
  • #9
Dick said:
It's not just an 'intuitive solution'. It's an informal description of a legitimate proof. How formal you want to make it depends on the requirements of your class. I think if you understand that and describe it clearly, that's all you need.

I was thinking of showing the same description for the proof. Usually, my professor is lenient but the people marking are very harsh and always demand a formal proof.
 
  • #10
SpringPhysics said:
I was thinking of showing the same description for the proof. Usually, my professor is lenient but the people marking are very harsh and always demand a formal proof.

I think you are being overly concerned. You can't prove the derivative of your integral is positive, because it's not. It does decrease sometimes but can never decrease to zero because of the area argument. Some things aren't worth being overly formal about.
 
  • #11
All right then, thanks to the both of you for your help!
 

What is the definition of a positive integral?

A positive integral is a whole number that is greater than zero. It is also known as a natural number, and can be represented on a number line starting from 1 and moving in increments of 1.

How can sums be used to prove a positive integral?

Sums can be used to prove a positive integral by showing that the sum of consecutive positive integers starting from 1 equals the given positive integral. This can be done through mathematical induction or by using the formula for the sum of an arithmetic sequence.

What is mathematical induction?

Mathematical induction is a method of proof used to show that a statement is true for all natural numbers. It involves proving a base case (usually n = 1) and then showing that if the statement is true for n, it is also true for n+1. This process is repeated until the statement has been proven for all natural numbers.

How can the sum of an arithmetic sequence be used to prove a positive integral?

The sum of an arithmetic sequence can be used to prove a positive integral by using the formula Sn = (n/2)(a + l), where Sn is the sum of the first n terms, a is the first term, and l is the last term. By substituting in the given positive integral for Sn and solving for n, it can be shown that the given positive integral is equal to the sum of n consecutive positive integers starting from 1.

What are some real-life applications of proving positive integrals using sums?

Proving positive integrals using sums is a fundamental concept in mathematics and has many real-life applications. It is used in counting and measuring, calculating distances and areas, and in various fields of science such as physics, chemistry, and engineering. It is also used in computer programming, where positive integrals are often used to represent quantities such as time, distance, and quantity of items.

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