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Proving Positive Integral using Sums 
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#1
Jan710, 05:34 PM

P: 107

1. The problem statement, all variables and given/known data
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.) 2. Relevant 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 3. 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. 


#2
Jan710, 07:25 PM

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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
Jan710, 07:30 PM

P: 107




#4
Jan710, 07:38 PM

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Proving Positive Integral using Sums
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
Jan710, 07:40 PM

P: 107

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 deltaepsilon 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). 


#6
Jan710, 09:10 PM

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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
Jan710, 09:26 PM

P: 107

Also, we have not learned how to differentiate an integral. 


#8
Jan710, 09:34 PM

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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
Jan710, 09:40 PM

P: 107




#10
Jan710, 09:55 PM

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#11
Jan710, 10:05 PM

P: 107

All right then, thanks to the both of you for your help!



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