The Problem:
Find I=∫02(x2+1)d⌊x⌋
My Doubt:
There seem two probable approaches to the question but both of them yield different results. Both seem to be mathematically correct and hence the confusion.
Approach 1:
We know that,
g(x).f′(x)+f(x).g′(x)=dxd(f(x).g(x))
Integrating the above expression we get:
∫abg(x)df(x)+∫abf(x)dg(x)=f(b).g(b)−f(a).g(a)
Using the above property to solve the integration we get the answer as 7
Approach 2:
We can write
I=∫01(x2+1)d⌊x⌋+∫12(x2+1)d⌊x⌋
But ⌊x⌋ assumes constant values of ⌊x⌋=0 and ⌊x⌋=1 in the respective intervals. And hence in both cases d⌊x⌋=0 and therefore I=0
I genuinely can't understand what's the correct method. I will really be grateful if someone can explain it to me.
Thank You!
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Top NewestFirst off, integration by parts doesn't work as the step function isn't differentiable.
Secondly, the integral is 2. I'll provide a proof of this later.
Btw, @Aditya Kumar The integral makes perfect sense and is defined.
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It would be really nice of you to put the proof here if possible! But I would also like to know then what's wrong with the Approach 1&2 illustrated above? It's a bit baffling
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Wait, the integral is 7.
We shall use the definition of integration as given in this book, with α(x)=⌊x⌋ and f(x)=1+x2.
Consider a partition P of [0,2] such that xr−1≤1≤xr where strict inequality holds in atleast one of the two inequalities. Also, by the definition of a partition, we have xn−1<2=xn. Then,
U(P,f,α)=f(xr)+f(xn)L(P,f,α)=f(xr−1)+f(xn−1)⟹infU(P,f,α)=supL(P,f,α)=f(1)+f(2)⟹∫02fdα=7
Note: I've only used the fact that f is monotonic.
Plus, in the book mentioned above, in the exercises, a more generalised version of IBP, valid for even discontinuos functions is given. You've made use of this version in approach 1 (luckily, the step function is monotonic).
If the integrals are calculated using the definition in approach 2, you get the right answer, 7.
@Miraj Shah @Aditya Kumar
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@Miraj Shah
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@Deeparaj Bhat ! Sorry for replying a bit late. Busy studying for the big one which is to be held on 22nd May! Actually the above problem was from a test paper it self! Had a doubt in the question, so thought of taking help from the brilliant Brilliant Community as I was pretty sure that I will be able to get at least some insight!
Oh yes! Really helpfulThank you!
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Your Approach 1 is fine.
In Approach 2, the floor function fails to be constant on the two subintervals; consider the two endpoints! The first integral takes the value of 1 due to the step at 1, and the second one takes a value of 5. With this correction, both approaches work and yield the same result.
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