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# Doubt in integration

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Note by M D
7 months, 2 weeks ago

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First things first - you have a function of $$x$$ as a limit of your integral. That wouldn't be a problem if your variable of integration wasn't also $$x$$; however, since that this is the case, I'm going to assume you meant to integrate up to $$\frac{ \lfloor y \rfloor }{n}$$ and make it a function of $$y$$.

Notice that $$f(x) = \{nx\}$$ is a periodic function with period $$\frac{1}{n}$$. This is because the fractional part of a number keeps repeating itself; another way to visualize this is to notice that:

$$f(x + \frac{1}{n}) = \{n*(x + \frac{1}{n})\} = \{1 + nx\} = \{nx\} = f(x)$$.

Given this, it shouldn't be hard to notice that $$a^{\{nx\}}$$ will also be a periodic funcion, since the exponent will keep repeating itself over and over and over. If we were to graph this, it would look something like this: http://imgur.com/a/4g17l

Now, notice that there are a certain number of slices to integrate, that are exactly equal sans their positioning. How many of those slices we have? Given the limits of the integral, we have exactly $$\lfloor y \rfloor$$ units of size $$\frac{1}{n}$$, so your integral becomes:

$$\int_{0}^{\frac{ \lfloor y \rfloor }{n}} a^{\{nx\}} dx = \lfloor y \rfloor * \int_{0}^{\frac{1}{n}} a^{\{nx\}} dx$$

At this point, we can remove the curly braces from the exponent since we are now dealing with the interval between 0 and $$\frac{1}{n}$$. This gives us:

$$\lfloor y \rfloor * \int_{0}^{\frac{1}{n}} a^{nx} dx$$.

Finally, integrating this assuming $$n$$ is constant, we get:

$$\lfloor y \rfloor * \int_{0}^{\frac{1}{n}} a^{nx} dx = \lfloor y \rfloor * \frac{1}{ln(a^{n})} * (a^{nx})\Big|_0^\frac{1}{n} = \frac{a - 1}{ln(a^{n})} * \lfloor y \rfloor$$

I hope this helps. If something I said wasn't clear, feel free to ask me. · 7 months, 2 weeks ago

Yes, if you take the limit to be $$\dfrac{ \lfloor y \rfloor }{n}$$, then your answer is perfectly right.

But the actual problem is this :

$$\displaystyle \int_0^{\frac{ \lfloor x \rfloor}{3}} \dfrac{27^{x}}{3^{ \lfloor 3x \rfloor}} dx$$

-Thanks · 7 months, 1 week ago

@M D You see, M, it doesn't make any sense to have the limit of integration be a function of x given that the variable of integration is x itself. Is this problem written in a textbook or test, and if that's the case would you mind uploading a picture of it? · 7 months, 1 week ago