How to calculate the integral of $f(x)$?

Question

Let $f(x)$ be a function which satisfies the following two properties:

1) $f(x) + f(-x) =2$

2) $f(1-x) = f(1+x)$

I need to calculate the $\int_0^{2016} f(x)dx$. I already tried to find $f(x)$ explicitly, but failed to do it.

I don't know how to solve it, what should I do? thanks

please show what you tried ! – alkabary May 27 '15 at 07:06 — alkabary, May 27 '15 at 07:06
are the f's in a and b the same? – Gappy Hilmore May 27 '15 at 07:06 — Gappy Hilmore, May 27 '15 at 07:06

score 5 · Answer 1 · answered May 27 '15 at 07:16

5

Hint: If the two $f$'s in (a) and (b) are same then, note that $$f(x)+f(2+x)=2\ \forall x\implies \int_{2k}^{2k+2}f(x) dx+\int_{2k+2}^{2k+4}f(x)dx=4\ \forall k$$ I hope you can take it from here.

answered May 27 '15 at 07:16

Samrat Mukhopadhyay

16,277

Hey I dont know how to take it from here... help please – Rotem ben May 27 '15 at 07:29
Now that you have $\int_{2k}^{2k+2}f(x) dx+\int_{2k+2}^{2k+4}f(x)dx=4\ \forall k$, you can write the integral $\int_0 ^{2016} f dx$ as $\sum_{k=1}^{1006} (\int_{2k}^{2k+2}f(x) dx+\int_{2k+2}^{2k+4}f(x)dx)=4\times 1006$ – Samrat Mukhopadhyay May 27 '15 at 07:32
but how can I use it here? I dont have X+2 or something.. – Rotem ben May 27 '15 at 07:36
Note that $f(-x)=f(1-(1+x))=f(1+1+x)=f(x+2)$, is it clear now? – Samrat Mukhopadhyay May 27 '15 at 07:37
how is it 1+1+x i didnt understand ... – Rotem ben May 27 '15 at 07:40
I used the condition $f(1-x)=f(1+x)$ in (b) – Samrat Mukhopadhyay May 27 '15 at 07:41
ohh sorry.. ok got it .. but how can I use your integrals?? – Rotem ben May 27 '15 at 07:43
Observe that $\int_{0}^{2016} f(x) dx=(\int_0^2 f(x) dx+\int_{2}^4 f(x) dx)+(\int_4^6 f(x) dx+\int_6^8 f(x) dx)+\cdots+(\int_{2012}^{2014}f(x) dx+\int_{2014}^{2016}f(x) dx)$. Do you see now? – Samrat Mukhopadhyay May 27 '15 at 07:45
so is it 2*(2016/4)?? – Rotem ben May 27 '15 at 07:47
It is $4*(2012)/4=2012$. – Samrat Mukhopadhyay May 27 '15 at 07:49
may you explain it again please? in one message? I dont understand – Rotem ben May 27 '15 at 07:49
Did you understand how I am breaking the integral in groups of two and then using the relation I stated in the answer? I f you have understood that then it only remains to count how many such groups there are. Clearly, the number of such groups is $2012/4$ as you can check. – Samrat Mukhopadhyay May 27 '15 at 07:51
why 2012/4 and not 2016/4? – Rotem ben May 27 '15 at 07:56
its not group of 4 from 2016? @Samrat Mukhopadhyay – Rotem ben May 27 '15 at 08:24
Just look at the first element of each group. The integration at the first element starts from $4k$, i.e. the lower limits are $0,4,8,12,\cdots,\ 2012$. Get my point? – Samrat Mukhopadhyay May 27 '15 at 08:50
Yea.. so the answer is 4*(2012/4)?? – Rotem ben May 27 '15 at 09:04
Exactly! Rotem ben. – Samrat Mukhopadhyay May 27 '15 at 09:41

How to calculate the integral of $f(x)$?

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