derivative of an integral

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Vali

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\(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R} ;f(a)=\int_{0}^{1}|x-a|dx\)
\(\displaystyle f'(2)=?\)
First exercise was to find f(2) which is 3/2.
How to derive an integral?
If f(2)=3/2 then f'(2) should be 0 but it's not.It's 1 the correct answer but I don't know why.What's method?
 
I would observe that for \(1<a\) we may write:

[MATH]f(a)=\int_0^1 a-x\,dx=a-\frac{1}{2}\implies f'(a)=1[/MATH]
 
Vali said:
If f(2)=3/2 then f'(2) should be 0...
Click to expand...

This is blatantly untrue. Just as one example, consider the function \(f(x) = x^2\). Then we have \(f(3) = 3^2 = 9\), which is a constant. But \(f^{\prime}(x) = 2x\), so \(f^{\prime}(3) = 2(3) = 6 \ne 0\).

In this case, the value of \(f(a)\) would appear to depend on both \(x\) and \(a\), but it doesn't really. The \(x\) gets "canceled" out during the evaluation of the integral, leaving only a function of \(a\). Since \(f\) is purely a function of \(a\), then its value at any point will always be a constant. Now, it's true that the derivative of a constant is 0... but that doesn't necessarily tell you anything about the value of \(f^{\prime}(a)\) at any given point. By your line of reasoning, every pure function of \(a\) would have a derivative of 0 at every point, because the value of \(f(a)\) is constant at every point.
 
Vali said:
\(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R} ;f(a)=\int_{0}^{1}|x-a|dx\)
\(\displaystyle f'(2)=?\)
First exercise was to find f(2) which is 3/2.
How to derive an integral?
If f(2)=3/2 then f'(2) should be 0 but it's not.It's 1 the correct answer but I don't know why.What's method?
Click to expand...
Are you saying that f'(2)=0 because f(2) = 3/2. That is because f(2) is a constant and the derivative of a constant is 0? If yes, then you really have to think about that.
Assuming 2 is in the domain of f and f: R-->R, then f(2) will be a real number for every function f. So are you saying that for all functions from R to R that has 2 in the domain f'(2) = 0? Even if the function is not differentiable at 2? Is this true for x=3? How about for x=-7? And my favorite, how about at x = ee?
Do you get what I am saying?
 
Vali said:
\(\displaystyle f:\mathbb{R}\rightarrow \mathbb{R} ;f(a)=\int_{0}^{1}|x-a|dx\)
\(\displaystyle f'(2)=?\)
First exercise was to find f(2) which is 3/2.
How to derive an integral?
If f(2)=3/2 then f'(2) should be 0 but it's not.It's 1 the correct answer but I don't know why.What's method?
Click to expand...
\(\displaystyle f(a) = \int_0^1 {|x - a|dx} =\begin{cases}\frac{1}{2}-a &: a\le 0 \\ a-\frac{1}{2} &: a\ge 1\\a^2-a+\frac{1}{2} &: \text{ elsewise }\end{cases}\)
Look here.
 
Yes, for \(0<a<1\) we have:

[MATH]f(a)=\int_0^a a-x\,dx+\int_a^1 x-a\,dx=\frac{1}{2}a^2+\frac{1}{2}-a-\frac{1}{2}a^2+a^2=a^2-a+\frac{1}{2}[/MATH]
 
Yes, I understood my mistake.I expressed myself in a wrong way.
Thank you for your help :)
 
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