Value of the integral using known conditions.

Question

Let there be a function on R such that $f (2-x)=f (2+x),f (4-x)=f (4+x) $ also $\int _0 ^2 f (x)dx=5$ then find $\int _{10} ^{50} f (x)dx $ . $$\text {Attempt} $$ $\int _0 ^2 f (x)=\int _0 ^2 f (2-x) $ ( From property). From given condition the integral equals $\int _0 ^2 f (2+x) $ again applying the property we get $\int _0 ^2 f (4-x) $ but this is not leading me anywhere. Also I think $f$ is symmetric about $x=2,4$ . But I dont know how thats possible. Hoping for hints and suggestions. Thanks

Answer 1

First, $f$ is periodic since

$$f(4+x) = f(4-x) = f(2+2-x) = f(2-(2-x)) = f(x)\text.$$

Second,

$$\int_2^4f(x)dx = \int_0^2f(x+2)dx = \int_0^2f(2-x)dx = -\int_2^0f(y)dy = \int_0^2f(x)dx\text.$$

From there, $$\int_{10}^{50}f(x)dx = \sum_{k=0}^{19}\int_{10+2k}^{10+2k+2}f(x)dx=20\cdot\int_0^2f(x)dx=100$$