Integration problem using inscribed rectangles

  • #1
chwala
Gold Member
2,668
352
Homework Statement
see attached
Relevant Equations
Integration -Hyperbolic Functions
Just went through this...steps pretty clear. I refreshed on Riemann integrals { sum of rectangles approximate area under curves}. My question is on the highlighted part in Red. The approximation of area under curve may be smaller or larger than the actual value. Thus the inequality may be ##<## or ##>##. Correct?

In the case of this question they chose ##>##. My point is they could have as well chosen to use ##<## with no implications.

1713046275919.png


1713046303668.png



My general comments on the ms approach is as follows,

##\cosh^{-1} r = \ln (r + \sqrt {r^2 -1} )## stems from one understanding the following steps

Let ##y = \cosh^{-1} x##

then

##x = \cosh y = \dfrac {e^y + e^{-y}}{2}##

##x= \dfrac{e^y + e^{-y}}{2}##

##2xe^y = e^{2y}+1##

...
##⇒ e^y = x ± \ln (x + \sqrt {x^2 -1} )##

##y = x ± \ln (x + \sqrt {x^2 -1} )##

They also used integration by parts noting that

##u = \cosh^{-1} x##

using
## \cosh^2 y - \sinh^2y = 1## and letting ## y =\cosh^{-1} x## then ##x = \cosh y##

##\dfrac{dy}{dx} = \dfrac{1}{\sinh y}##

##\sinh y = \sqrt{\cosh^2y -1}##

##\dfrac{dy}{dx}= \dfrac{1}{\sqrt{x^2-1)}}##

and lastly,


##\ln (r + \sqrt {r^2 -1} ) > [x \ln r + \sqrt {r^2-1}]_1^N - [\sqrt {x^2-1}]_1^N##

##\ln (r + \sqrt {r^2 -1} ) > N \ln N +\sqrt {N^2-1} -0-\sqrt {N^2-1}+0##

##\ln (r + \sqrt {r^2 -1} ) > N \ln N +\sqrt {N^2-1}-\sqrt {N^2-1}+0##
 
Last edited:
Physics news on Phys.org
  • #2
chwala said:
The approximation of area under curve may be smaller or larger than the actual value. Thus the inequality may be ##<## or ##>##. Correct?
The correct inequality is
$$\sum_{i=2}^N\cosh ^{-1}i>\int^N_1 \cosh ^{-1}x dx$$
because the Riemann sum uses the right endpoints of each subinterval and the function is increasing. The inequality could be wrong depending on if it's a left, middle, or trapezoidal Riemann sum.
 
  • Informative
  • Like
Likes SammyS and chwala
  • #3
For a general Riemann sum is evaluated over the interval ##[a,b]##, if we evaluate the sum using increasingly narrower rectangles, the limit as the width of the rectangles go to zero is exact integral. Specifically using the right endpoint rule, with partition ##a=x_1<\cdots <x_n=b##,
$$\lim_{n\to \infty}\sum^{n}_{i=1}f(x_{i+1})(x_{i+1}-x_i)=\int^b_af(x)dx.$$
 
  • Like
Likes chwala
  • #4
The above is the meaning/definition of Riemann integrability. The convergence in the sense that as ##Lim ||P||_{\rightarrow 0}##, meaning the maximum width of the rectangles becomes indefinitely small, the value of the Riemann Integral becomes indefinitely -close to a value ##R## ; in a" ## P-\epsilon##" sense, of the Riemann integral. This is called Net Convergence.
 
  • Informative
  • Like
Likes docnet and chwala
  • #5
WWGD said:
The above is the meaning/definition of Riemann integrability. The convergence in the sense that as ##Lim ||P||_{\rightarrow 0}##, meaning the maximum width of the rectangles becomes indefinitely small, the value of the Riemann Integral becomes indefinitely -close to a value ##R## ; in a" ## P-\epsilon##" sense, of the Riemann integral. This is called Net Convergence.
Just want to say, I noticed that my comment isn't right. I stated the limit as the number of rectangles go to infinity, but I didn't specify that the width of every rectangle goes to zero, or equivalently that ##P=\text{max}_{i\in I}\left(x_{i+1}-x_i\right),## ##I=\{1,...,N\}## goes to zero. I had a feeling that something was off, but didn't notice it right away.

Does the '##P - \epsilon##' sense of convergence mean that, for every ##\epsilon>0##, there exists a ##P## such that the corresponding Riemann sum and the value ##R## differ by ##\epsilon##?
 
  • #6
docnet said:
Just want to say, I noticed that my comment isn't right. I stated the limit as the number of rectangles go to infinity, but I didn't specify that the width of every rectangle goes to zero, or equivalently that ##P=\text{max}_{i\in I}\left(x_{i+1}-x_i\right),## ##I=\{1,...,N\}## goes to zero. I had a feeling that something was off, but didn't notice it right away.

Does the '##P - \epsilon##' sense of convergence mean that, for every ##\epsilon>0##, there exists a ##P## such that the corresponding Riemann sum and the value ##R## differ by ##\epsilon##?
Let ##R(f)## be the Riemann sum associated with ##f##.Then the Riemann Integral of ##f## converges to ##R## means that for every ##\epsilon >0##, there is a## \delta## , so that ##|R(f) -R|<\epsilon## for any partition ##P## with ##||P||<\delta##.
 
  • Like
Likes docnet
  • #7
This isn't a question on Riemann sums; it's a question on the integral comparison test for the convergence of a series.
 
  • #8
pasmith said:
This isn't a question on Riemann sums; it's a question on the integral comparison test for the convergence of a series.
I'm answering a question that was asked.
 
  • #9
WWGD said:
I'm answering a question that was asked.

I'm commenting on the OP's choice of thread title.
 
  • #10
pasmith said:
I'm commenting on the OP's choice of thread title.
Ok, my bad.
 
  • #11
pasmith said:
I'm commenting on the OP's choice of thread title.
I changed the thread title to "Integration problem using inscribed rectangles." I hope this better describes what the thread is about. Possibly @pasmith's recommendation of "integral comparison test" might be a better choice, though.
 
  • #12
Mark44 said:
I changed the thread title to "Integration problem using inscribed rectangles." I hope this better describes what the thread is about. Possibly @pasmith's recommendation of "integral comparison test" might be a better choice, though.
Then we may have to go through several likes of " proof this solution" , " Avance Calculas" I've seen. Should I point them out to staff?
 
  • #13
WWGD said:
Then we may have to go through several likes of " proof this solution" , " Avance Calculas" I've seen. Should I point them out to staff?
I regularly edit thread titles with gross typos, but haven't seen any lately like the ones you cited. @berkeman also changes thread titles if they don't match the thread content very well or aren't descriptive.
If you see any that you think ought to be changed, of relatively recent threads, yes, go ahead and report them.
 
  • Like
Likes berkeman
  • #14
WWGD said:
Let ##R(f)## be the Riemann sum associated with ##f##.Then the Riemann Integral of ##f## converges to ##R## means that for every ##\epsilon >0##, there is a## \delta## , so that ##|R(f) -R|<\epsilon## for any partition ##P## with ##||P||<\delta##.
If I understand correctly, then ##||P||## means $$\sqrt{\sum_i\left(\Delta x_i^2\right)}?$$ Thanks
 
  • #15
docnet said:
If I understand correctly, then ##||P||## means $$\sqrt{\sum_i\left(\Delta x_i^2\right)}?$$ Thanks

[itex]\|P\| = \max_i \Delta_i[/itex].
 
  • Like
Likes docnet

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
785
  • Calculus and Beyond Homework Help
Replies
2
Views
497
  • Calculus and Beyond Homework Help
Replies
10
Views
539
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
342
  • Calculus and Beyond Homework Help
Replies
2
Views
549
  • Calculus and Beyond Homework Help
Replies
1
Views
529
  • Calculus and Beyond Homework Help
Replies
11
Views
455
  • Calculus and Beyond Homework Help
Replies
20
Views
500
  • Calculus and Beyond Homework Help
Replies
2
Views
554
Back
Top