If $f(x) + f(x + 1) = 2$, $I_1 = \int^8_0 f(x)dx$ and $I_2 = \int^3_{-1} f(x) dx$ then find $I_1 + 2I_2$.

Question

Given

$f(x) + f(x + 1) = 2$ (f is continuous)

$I_1 = \int^8_0 f(x)dx$

$I_2 = \int^3_{-1} f(x) dx$

Find $I_1 + 2I_2$.

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I found $f(x) = f(x+2)$ so function has a period of 2.

So $I_1 = 4\int^2_0 f(x)dx$ but I'm not getting how to solve this.

Answer 1

$I_2 = \int^3_{-1} f(x) dx$, let $t=x+1$, then we have $I_2 = \int^4_{0} f(t-1) dt$, where $f(t-1)=2-f(t)$,

$$I_2 = \int^4_{0} 2-f(t) dt=8-\int^4_{0} f(t) dt\tag{1}$$

Since the function has a period of $2$, so we have $f(x)=f(x+2)=f(x+4)$,

$$I_1 = \int^8_0 f(x)dx=\int^4_0 f(x)dx+\int^8_4 f(x)dx$$

Let $t=x-4$, then the second integral becomes

$$\int^8_4 f(x)dx=\int^4_0 f(t+4)dt=\int^4_0 f(t)dt$$

Therefore,

$$I_1 = 2\int^4_0 f(x)dx \tag{2}$$

Combine $(1)$ and $(2)$

$$I_1+2I_2=16$$