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:

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.
–
Alexandre Miquilino
·
11 months, 1 week ago

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@Alexandre Miquilino
–
Yes, if you take the limit to be \(\dfrac{ \lfloor y \rfloor }{n} \), then your answer is perfectly right.

@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?
–
Alexandre Miquilino
·
11 months, 1 week ago

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@Alexandre Miquilino
–
This was given as an assignment. You are right. May be I noted down the problem wrong.
Thanks anyways :)
–
M D
·
11 months, 1 week ago

## Comments

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TopNewestFirst 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. – Alexandre Miquilino · 11 months, 1 week ago

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But the actual problem is this :

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

-Thanks – M D · 11 months, 1 week ago

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– Alexandre Miquilino · 11 months, 1 week ago

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?Log in to reply

– M D · 11 months, 1 week ago

This was given as an assignment. You are right. May be I noted down the problem wrong. Thanks anyways :)Log in to reply