$g(2-x)=g(2+x)$ and $g(4-x) = g(4+x)$ and $ \int^{2}_{0}g(x)dx = 2.$ then $\int^{50}_{0}g(x)dx$ is

Question

function $g(x)$satisfy following condition $g(2-x)=g(2+x)$ and $g(4-x) = g(4+x)$

and $\displaystyle \int^{2}_{0}g(x)dx = 2.$ then $\displaystyle \int^{50}_{0}g(x)dx$ is

from $g(2-x) = g(2+x)$

$g(2-(2-x)) = g(2+2-x)$

$g(x) = g(4+x)=g(4-x)=g(x+4)$

time period of $g(x) = 4$

i wan,t go after that, could some help me with this

Hint: Note that $\int_2^4g(x)dx=\int_0^2g(x+2)dx=\int_0^2g(2-x)dx$ Now let $u=2-x$ — lulu, Dec 20 '16 at 19:09

score 2 · Accepted Answer · answered Dec 21 '16 at 11:29

In fact, you already know that $g(x)$ is a periodic function, and $T=4$, the image of $g(x)$ is symmetric about x = 2.

So,we have:$$\int_0^4g(x)\,dx=\int_0^2g(x)\,dx+\int_2^4g(x)\,dx$$ $$=2\int_0^2g(x)\,dx=4$$

then,$$\int_0^{50}g(x)=12\int_0^4g(x)\,dx+\int_{48}^{50}g(x)\,dx$$ $$=48+\int_0^2g(x)\,dx=50$$

PS:If you use the function of graphics, you will be better understood

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