Solving definite integrals with periodic integrand

Question

So, the question is: If $f(2-x) = f(2+x)$ and $f(4-x) = f(4+x)$ where $f(x)$ is a function for which $$\int_{0}^{2}f(x)dx =5$$ ,

then prove that : $$\int_{0}^{50}f(x)dx =125$$ $$\int_{-4}^{46}f(x)dx =125$$ $$\int_{2}^{52}f(x)dx =125$$

also comment whether $\int_{1}^{51}f(x)dx =125$ is true or false.

My approach: since $f(2-x) = f(2+x)$ , replace $x$ by $x+2$ and we get $f(x) = f(4+x)$ hence period of $f(x)$ is $4$ now,

$$\int_{0}^{50}f(x)dx = \int_{0}^{48}f(x)dx +\int_{0}^{2}f(x)dx $$ hence $$\int_{0}^{50}f(x)dx =12* \int_{0}^{4}f(x)dx +\int_{0}^{2}f(x)dx $$ which becomes $$\int_{0}^{50}f(x)dx =12*( \int_{0}^{2}f(x)dx+\int_{0}^{2}f(x+4)dx) +5 $$ so,we get $$\int_{0}^{50}f(x)dx =125$$

similarly for $$\int_{-4}^{46}f(x)dx $$,substitute $x+4=t$ therefore, $$\int_{-4}^{46}f(x)dx =\int_{0}^{50}f(t)dt =125 $$

however I am not able to prove on the final part and comment on whether $\int_{1}^{51}f(x)dx =125$ is true or false.Kindly help me out.

Answer 1

Hint: $\int_2^{52} f(x)dx=\int_0^{52} f(x)dx -\int_0^{2} f(x)dx=(13)\int_0^{4} f(x)dx-5$.

Answer 2

$$\begin{align} \int_{1}^{51} f(x) dx &=\int_{1}^{3} f(x) dx+\int_{3}^{51} f(x) dx\\ &=\int_{0}^{2} f(x+1) dx+120\\&\ne 125\tag{$\because\int_{0}^{2}f(x)dx=5$}\end{align}$$