Simplifying a strange integral

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This looks strange and hard, but with careful analysis it becomes simple to work out. Thanks Sundipan for the suggestion!

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fun fact : x - floor(x) gives the decimal part of x

noskillman
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Thats a famous variation of a jee mains question. Since I have to appear for that exam on 8th, it was a nice revision. Thanks

ttyagraj
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I swear Engineering school ruins people. I see these problems and my instinct is *always* the brute force method. 100 repeating intervals? Find the area of each and sum them up, could probably write an algorithm to do that. And then seeing the graceful, simple solution is always a brick to the face, seeing that it can be done in like three lines of math. Absolutely wild.

mrsqueaksrules
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whenever you see floor function, partition the interval of the integral into (k, k+1) subinterval form.

= (₀∫¹+₁∫²+₃∫⁴+...+₉₉∫¹⁰⁰) e^(x–⌊x⌋) dx
= ₙ₌₀∑⁹⁹ ₙ∫ⁿ⁺¹ e^(x–⌊x⌋) dx
= ₙ₌₀∑⁹⁹ ₙ∫ⁿ⁺¹ eˣ⁻ⁿ dx
= ₙ₌₀∑⁹⁹ eˣ⁻ⁿ ₙ|ⁿ⁺¹
= ₙ₌₀∑⁹⁹ (eⁿ⁺¹⁻ⁿ–eⁿ⁻ⁿ)
= ₙ₌₀∑⁹⁹ (e–1)
= 100(e–1)

spiderjerusalem
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Wow it's awesome haven't studied calculas but I could still connect it

curddyhealer
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This was a delightful little problem! 🤗

kennethsizer
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I could do it in my head and its really very easy for a person preparing for JEE Advanced

rex_yourbud
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The mitochondria is the powerhouse of the cell.

jackkraken
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Ahh these functions with vertices - dude its so satisfying to watch how the value literally "oscillates" from a variable to a constant and vice versa, according to the interval in which it's integrateddd 😊

sumitkundu
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Now for Jee advanced aspirants this question is likely to solve within few seconds in mind

Subhalin
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floor(x) = Greatest Integer Function [•]

orenawaerenyeager
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I had no trouble solving this one, but I don't think it's particularly nice or intellectually pleasing.

Incidentally, that repeating function isn't drawn correctly as it appears to be a straight line, whilst it's actually exponential. It makes it look like it can be solved geometrically.

TheEulerID
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Interesting problem! Now I'm trying to think of a use case 🤔. Maybe estimating compounded errors for a a finite resolution motor or other actuator.

matthewjohnstone
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Or exp(floor(a)-a) -1 + (e-1)floor(a) if you integrate up to a general positive real number a instead of 100

joefarrow
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I tried doing it in my head, so here’s my answer beforehand: 100(e-1)

Edit: Yippee!

newmanhiding
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Me as jee aspirant solves without pen 😂😂

ravikumar-hjti
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A Very common math problem in almost all Indian exams

g_sreemanpattanaik
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The function that is integrated must be continuous to be evaluated, right ?

ΧρήστοςΤριανταφυλλίδης-βγ
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There are 3 things I have no idea what they are
Integral
Floor of X
And dx

globalwarrior
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I failed calc 2 twice why am I even watching these

elizabethstauffer