73 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.my obtained integral. not able to solve it
This seems to be a question of integral comparisions.
Right. That makes sense
Can u pls tell 5th too
Sorry, I just realized I didn't tell how. ðŸ˜
Did you get the answer?
$\frac{1}{1+x^{2}.e^{x}} \leq 1$ in [0,1]
SirLancelotDuLac
Is it in another thread?
Yeah I just found f(0) and f(1). Fs the integral would lie between them so gif is computable
Mb
Ah, I still can't see the question though :/
Upper reimann + lower reimann / 2* integral ka gif
This module might be goated ðŸ˜
I believe this is (midpoint riemann)/integral or basically integral and midpoint riemann sum ka comparision, which I need to ponder over.
Although @Opt sorry to disturb can you check this out?
Also correction, it's trapezoidal riemann not midpoint riemann.
Woah I have been summoned
@hardcoreisdead is it zero?
Integral comparisons?
@SirLancelotDuLac pls verify
Woah
Insane stuff
Invictus ki module kaha se khareedu
Legit this has beautiful questions
Ignore the bad handwriting btw
I was writing while having coffee
Is this question based on concepts of definite integration or does itninvolve basic concepts
Also, tbh you only need the upper bound of π/4, since the integral is definitely positive (non-negative function at all points), but i did the lower bound first so it stuck lol
Basic stuff
What does riemann mean?
Riemann was a mathematician. Who did a lot of work on rigorously defining the definite integral. And his formulation of it involves taking the limit of a sum of approximate areas underneath pieces of curves to compute the integral. This summation is usually called a Riemann sum, and depending on the ranges, it can be different kinds of Riemann sum.
The normal integral we learn is called a Riemann integral because we use his definition of things
Ohh
If instead you want to integrate certain weird functions with infinite (dense) discontinuities and stuff, Riemann integrals break down. Don't work.
You need a different thing then called Lebesgue integrals
Not in syllabus
So chhodo
Okayy
Now lemme see the question
Also can u check integration 2 thread please
Small mistake here. The sign got mixed up in the 1/1+x²exp(x) and 1/1+x² part. Mb
Ping me man
@Opt Lancelot wanted u to look at 5
Not that one, the question below.
Basically how can we compare trapezoidal riemann sum and integral of a function.
Oh
Mb
Wait there's no limit there
Hmmm
@hardcoreisdead I'm not sure but I somehow got 1
1
I got 1
Which seems much more likely
I just did the easiest thing which is to plug in a function and check
The ans is zero
Oh
Huh
OH
Omg I understand why
When you approximate with trapezoids, the trapezoids are always under the convex function.
Area of trapezoids is strictly lesser than the area under function
Therefore the ratio must be less than one
In both increasing and decreasing functions ???
Ohhh
Ye I missed this.
Depending on concavity. If it was concave up it the answer would have been 1.
Yeah, that's why the second derivative is specified negative here
Yepp.
(Also sowwy to invoke you out of the blue ;_;)
No prob I'm bored out of my mind anyways
And jobless
im still shocked that by April-May next year ill be official categorised as unemployed
@Opt u free??
I need to clear some doubts regarding this
Sure
Go on
All this is correct ??
It probably is
I wanna know how the selected area gives the upper and lower bounds of the summation
Like why r=0 to n-1 in the first case
Kahan gye ....
Yeah wait
Was talking in another server
Alr
Write it out termwise
And you'll see how the height of the rectangles varies
Samjha nahi.....
Oh.
Like this???
Achaa
Acha so the graphs I sent above proves these two statements
Now how do I relate this result with the trapezoidal result u told
Basically how do I prove this ....
@Opt
@SirLancelotDuLac this.
Try to think of it as trapezoids.
Isme consider the first figure but intead of squares, we are taking trapezoids, the area of which is (f(x)+f(x+dx))dx/2
After that you'll be able to see it.
Like this..
Can u share a better graph. Mine is 🤡
Also I just realised original ques mein , ques 5 , f"(x) < 0 dia hai . Shouldn't it be f'(x) <0 @SirLancelotDuLac
Nah only concavity we need
Ye this shows this. (Search up trapezoidal reimann sum)
So ismein f"(x) aayega f'(x) ki jagah ???
Nope.
Ye trapezoidal nahi hai na. Think of it this way: f(x+dx)>f(x) if f'(x)>0
Toh upar wali inequality makes sense, if f'(x)>0
So ismein f'(x) ki jagah f"(x) aayega. Because the trapezoid thing Is based on concavity
Yepp. Trapezoid mein concavity matters kyunki
Agar do points hai with diff. y coordinates
And usko ek seedhe line se jodo
Aur ek concave up aur concave down curve se jodo
And then you draw verticals from both the points to x axis
You can see how the area varies
But normal reimann sums mein the thing is, it is horizontal so if the line goes below horizontal level (where function is decreasing, nothing to do with concavity) the area becomes lesser.
i still dont quite understand how the inequality obtained from this rule transltes to that summation form
Consider x=r/n and x=(r+1)/n, unn dono ke trapezoid ka area ka summation is 1/n[f(r/n)+f((r+1)/n)]/2 ka summation
Which is 0.5(summation of 1/n.f(r) and summation of 1/n.f((r+1)/n))
Where r goes from 0 to n-1
Sub. r+1=t in the second summation
Perfect ???
Actual mathematical proof ke pov se batana
@SirLancelotDuLac
,rotate
Yes. (This would be for concave down graphs)
haan
every term , applicationof limits is correct right?
going from x2 x1 format to r/n and r+1/n
Ye man, this is correct.
alr
big thanx
+solved @SirLancelotDuLac @Opt
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