What is the inner product of two piecewise-defined functions?

Asuralm
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Dear all:
I have a problem about the inner product of a function. Give a function

<br /> \begin{displaymath}<br /> f(x) = \left\{ \begin{array}{ll}<br /> x &amp; \textrm{if $x \in [0,1]$}\\<br /> -x+2 &amp; \textrm{if $x \in (1, 2]$}<br /> \end{array}<br /> \end{displaymath}<br /> \{

What's the value of the inner product of the function itself over [0,2]?
<br /> \begin{displaymath}<br /> &lt;f(x), f(x)&gt; = \int_{x=0}^{x=2} f(x)f(x) d_x<br /> \end{displaymath}<br />]

If given another function
<br /> <br /> g(x) = \left\{ \begin{array}{ll}<br /> x-1 &amp; \textrm{if $x \in [1,2]$}\\<br /> -x+3 &amp; \textrm{if $x \in (2, 3]$}<br /> \end{array}<br /> <br /> \{

What's the inner product of f(x) and g(x) please?

Thanks for answering.
 
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For you first question you have to separate integral into two
One of them is from 0 to 1, the other is from 1 to 2.

For the second you have to explain on which interval we take the inner product they are from different worlds.
 
I know the principle actually. Could you give me the whole details please? Because I can't get the correct answer.
 
For question1
You have to get from integral(0-1) =1/2 and from integral(1-2) =1/3
If you did not then write what you did .Maybe we can find the mistake
It would be yours or mine
 
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for question 2 : I am still waiting an explanation
It can be only defined on [1,2] i think
 
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It's possible that the intention is that f and g vanish wherever not explicitly defined. Then you would be right, it would be like on [1,2]...
 

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