Let f : [ 0 , 1 ] → R be a function defined as follows: f ( x ) = 1 0 x sin x + cos x
Enter the value of the following expression:
⌊ ∫ 0 1 f ( x ) d x ⌋
Details and Assumptions:
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There's a more simpler method to do this. You can approximate the integral. You'll get something lesser than 5.
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In the subjective exam, you don't have much time (I'm kinda slow in computations), plus if the method is non trivial, I'll have to provide a proof of it. This was what striked me first in the exam, so I went with it.
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Ooh I didn't know that was a subjective exam.
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@Aditya Kumar – There's no way to find the closed form of it right?
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@A Former Brilliant Member – By real analysis I don't think so we can get a closed form. But by complex analysis there might be one. @Mark Hennings can help us.
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@Aditya Kumar – Waiting for closed form.
@Aditya Kumar – I'm doubtful that a closed form can be found; I certainly don't know one. My solution to this problem was Deeparaj's...
You can use am-gm inequality also right?
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I'm not sure how you intend to achieve that. Could you post a solution that uses it?
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I can't post a solution write now as I do not have a laptop and I do not know latex. I'll try to post it as soon as possible. I think these problems might interest you,
https://brilliant.org/problems/check-this-out-3/?ref_id=1271985
https://brilliant.org/problems/convergency-and-divergency/?ref_id=1198750
https://brilliant.org/problems/think-this-is-easy/?ref_id=1260048
If u liked them I think you are welcome to try my set,
https://brilliant.org/profile/abhi-pwu19k/sets/my-creations-check-them-out/?ref_id=1198669
Thank you!!
Perfect solution!! Upvoted!!
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We note that sin x + cos x = 2 cos ( x − 4 π ) ∈ [ 1 , 2 ] ∀ x ∈ [ 0 , 1 ]
Hence, as x ∈ [ 0 , 1 ] , 1 0 x 2 ≤ f ( x ) ≤ 1 0 x With equality holding if and only if x = 0 o r 1 .
As equality holds only for finitely many points, the inequalities become on strict integrating on all sides.
Hence, 4 < 1 0 ( 2 − 1 ) < ∫ 0 1 f ( x ) d x < 5 ⟹ ⌊ ∫ 0 1 f ( x ) d x ⌋ = 4