A Calculus Problem!

The Problem:

Find I=02(x2+1)dxI=\displaystyle\int_{0}^{2} (x^2+1)\,d\left \lfloor x\right \rfloor

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)=d(f(x).g(x))dxg(x). f'(x) + f(x). g'(x) = \large \frac{d\left(f(x). g(x)\right)}{dx}

Integrating the above expression we get:

abg(x)df(x)+abf(x)dg(x)=f(b).g(b)f(a).g(a)\displaystyle\int_{a}^{b} g(x)\,df(x) + \displaystyle\int_{a}^{b} f(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\boxed{7}

Approach 2:

We can write

I=01(x2+1)dx+12(x2+1)dxI=\displaystyle\int_{0}^{1} (x^2+1)\,d\left \lfloor x\right \rfloor +\displaystyle\int_{1}^{2}(x^2+1)\,d\left \lfloor x\right \rfloor

But x\left \lfloor x\right \rfloor assumes constant values of x=0\left \lfloor x\right \rfloor=0 and x=1\left \lfloor x\right \rfloor=1 in the respective intervals. And hence in both cases dx=0d\left \lfloor x\right \rfloor=0 and therefore I=0I=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!

Note by Miraj Shah
5 years ago

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1 vote

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First 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.

Deeparaj Bhat - 5 years ago

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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

Miraj Shah - 5 years ago

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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\alpha(x)=\lfloor x \rfloor and f(x)=1+x2f(x)=1+x^2.

Consider a partition PP of [0,2][0,2] such that xr11xrx_{r-1}\leq 1\leq x_{r} where strict inequality holds in atleast one of the two inequalities. Also, by the definition of a partition, we have xn1<2=xnx_{n-1}<2=x_n. Then,

U(P,f,α)=f(xr)+f(xn)L(P,f,α)=f(xr1)+f(xn1)    infU(P,f,α)=supL(P,f,α)=f(1)+f(2)    02fdα=7U(P, f, \alpha)=f(x_r)+f(x_n) \\ L(P, f, \alpha)=f(x_{r-1})+f(x_{n-1}) \\\implies \inf U(P, f, \alpha)=\sup L(P, f, \alpha)=f(1)+f(2) \\\implies \int_0^2 f \, d\alpha = 7

Note: I've only used the fact that ff 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

Deeparaj Bhat - 5 years ago

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@Deeparaj Bhat Was this helpful? @Miraj Shah

Deeparaj Bhat - 5 years ago

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@Deeparaj Bhat Oh yes! Really helpful @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!

Thank you!

Miraj Shah - 5 years ago

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@Miraj Shah You're writing any other exams? (Except JEE ADVANCED, which is on 22nd)

Deeparaj Bhat - 5 years ago

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@Deeparaj Bhat Actually yes. BITS and all! You have to give JEE this year?

Miraj Shah - 5 years ago

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@Miraj Shah Yup. That, BITSAT and CMI in my case.

Deeparaj Bhat - 5 years ago

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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.

Otto Bretscher - 5 years ago

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