# definite integral with periodic function of period = 1

If $f(x)+f\left(x+\frac{1}{2}\right) = 1$. Then $\displaystyle \int_{0}^{2}f(x)dx =$

If function $f(x)$ is Continuous and Differentiable in $x\in \left(0,2\right)$

My Try:: Using $f(x)+f\left(x+\frac{1}{2}\right) = 1$

Replace $x$ by $\displaystyle x+\frac{1}{2}\;,$ We Get

$f\left(x+\frac{1}{2}\right)+f(x+1) = 1$ Using These two , We Get

$f(x+1) = f(x)$ , That means $f(x)$ is a periodic function with period $= 1$

So $\displaystyle \int_{0}^{2}f(x)dx = \int_{0}^{1}f(x)dx + \int_{1}^{2}f(x)dx$

In Second Integral Put $x = t+1$ and $dx = dt$ and changing limit , we get

$\displaystyle = \int_{0}^{1}f(x)dx+\int_{0}^{1}f(t+1)dt$

$\displaystyle = \int_{0}^{1}f(x)dx+\int_{0}^{1}f(x+1)dx$

$\displaystyle = \int_{0}^{1}f(x)dx+\int_{0}^{1}f(x)dx = 2\int_{0}^{1}f(x)dx$

Ans Given is $= 1$ but I did not understand how can i get it

-
$\int{f\left(\left(\text{upper limit} + \text{lower limit}\right) - x\right)}dx$ – hjpotter92 Jan 28 '13 at 17:37
Thanks Back in a Flesh – juantheron Jan 28 '13 at 17:44

If $f(x+\frac 1 2) + f(x) = 1$ then $$\int_a^{a+1}f(x) dx = \int_a^{a+\frac 12} f(x) dx + \int_{a+\frac 12}^{a+1} f(x) dx = \int_a^{a+\frac 12} \left[f(x) + f(x+ \frac 12)\right] dx = \frac 12$$

-
Thanks manu-fatto got it – juantheron Jan 28 '13 at 17:44

hint:integrate both side of the $$f(x)+f\left(x+\frac{1}{2}\right) = 1$$ let $u:=x+\frac12$ and you can solve it easily

-
Thanks Maisam Hedyelloo – juantheron Jan 28 '13 at 17:45
@ juantheron :your welcome . – Maisam Hedyelloo Jan 28 '13 at 17:46

You have $$b-a=\int_a^b 1\,dx=\int_a^bf(x)\,dx+\int_a^bf\Bigl(x+\frac12\Bigr)\,dx=\int_a^bf(x)\,dx+\int_{a+\frac12}^{b+\frac12}f(x)\,dx\,.$$ Now take $a=0, b=a+\frac12$, obtaining $\int_0^1f(x)\,dx=1/2$.

-
Thanks Matemáticos Chibchas – juantheron Jan 28 '13 at 17:49