• Feb 20th 2013, 05:01 AM
James7361539
Hey I'm stuck on this Q and I would be grateful for any replies.

I have tried twice to write the problem in latex but unfortunately I am not proficient enough in it.

I have attached the Q NB it is Q2 on the attached photo
Also I have attached the graph that I think is right for the first part
but my real Q is for part 1 how do you show that that integral is equal to that sum? I can see it clearly from
the graph but I don't know how to show it?
I'll work on part two later.

I would be really grateful for any replies.

Thanks from James
• Feb 20th 2013, 05:55 AM
Kmath
First,
$\int_{0}^a\sqrt{[x]}dx= \int_{0}^1\sqrt{[x]}dx+\int_{1}^2\sqrt{[x]}dx+\cdots +\int_{a-1}^a\sqrt{[x]}dx$

second, note that in the first interval (integral) $[x]=\sqrt{0}$
in the second interval (integral) $[x]=\sqrt{1}$
and so on, unitil
the ath integral $[x]=\sqrt{a-1}$, so
$\int_{0}^a\sqrt{[x]}dx= \int_{0}^10dx+\int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx$
$= \int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx =1+2+\cdots+\sqrt{a-1}$ as desired.
• Feb 20th 2013, 07:01 AM
James7361539
Quote:

Originally Posted by Kmath
First,
$\int_{0}^1\sqrt{[x]}dx= \int_{0}^a\sqrt{[x]}dx+\int_{1}^2\sqrt{[x]}dx+\cdots +\int_{a-1}^a\sqrt{[x]}dx$

second, note that in the first interval (integral) $[x]=\sqrt{0}$
in the second interval (integral) $[x]=\sqrt{1}$
and so on, unitil
the ath integral $[x]=\sqrt{a-1}$, so
$\int_{0}^1\sqrt{[x]}dx= \int_{0}^a0dx+\int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx$
$= \int_{1}^21dx+\cdots +\int_{a-1}^a\sqrt{a-1}dx =1+2+\cdots+\sqrt{a-1}$ as desired.

Hi I have no idea why the first part of your answer is true?
I don't really even understand what it means?
• Feb 20th 2013, 07:53 AM
Kmath
Quote:

Originally Posted by James7361539
Hi I have no idea why the first part of your answer is true?
I don't really even understand what it means?

you are right. I correct the bounds
• Feb 20th 2013, 09:31 AM
James7361539
Quote:

Originally Posted by Kmath
you are right. I correct the bounds

Hey Kmath

I now understand how to do part 1 but now I am stuck on part 2

I have attached the graph of y=2^[x] which I think will help.

I would be grateful if you could please explain how you do part 2

Thanks
• Feb 20th 2013, 09:53 AM
Kmath
Ok, again we have
$\int_0^a{2^{[x]}dx}=\int_0^1{2^{[x]}dx}+\int_1^2{2^{[x]}dx}+\cdots+\int_{a-1}^a{2^{[x]}dx}$
as the fisrt part we know the value of ${[x]}$ on each interval, that is
$\int_0^a{2^{[x]}dx}=\int_0^1{2^{0}dx}+\int_1^2{2^{1}dx}+\cdots+ \int_{a-1}^a{2^{a-1}dx}$
$=\int_0^1{1dx}+\int_1^2{2dx}+\cdots+\int_{a-1}^a{2^{a-1}dx}$
All the integrants are constants and the lenght of of each interval is 1 we have
$\int_0^a{2^{[x]}dx}=1+2+4+\cdots+{2^{a-1}}$
The later series is finite geometric series and its sum gives you the result
• Feb 22nd 2013, 01:07 AM
altoonje
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• Feb 22nd 2013, 03:12 AM
James7361539
Quote:

Originally Posted by Kmath
Ok, again we have
$\int_0^a{2^{[x]}dx}=\int_0^1{2^{[x]}dx}+\int_1^2{2^{[x]}dx}+\cdots+\int_{a-1}^a{2^{[x]}dx}$
as the fisrt part we know the value of ${[x]}$ on each interval, that is
$\int_0^a{2^{[x]}dx}=\int_0^1{2^{0}dx}+\int_1^2{2^{1}dx}+\cdots+ \int_{a-1}^a{2^{a-1}dx}$
$=\int_0^1{1dx}+\int_1^2{2dx}+\cdots+\int_{a-1}^a{2^{a-1}dx}$
All the integrants are constants and the lenght of of each interval is 1 we have
$\int_0^a{2^{[x]}dx}=1+2+4+\cdots+{2^{a-1}}$
The later series is finite geometric series and its sum gives you the result

Thanks for all your help Kmath.
I now understand how to do this Q