Animated Solution for Mathematics - Definite Integration: Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function such that $f(x)+f(x+1)=2$, for all $x\in \mathbb{R}$. If $I_1={\int }_0^{8}f(x)dx$ and $I_2={\int }_{-1}^{3}f(x)dx$, then the value of $I_1+2I_2$ is equal to ____

Welcome, future engineers! Today, we are going to unravel a problem that might look like a standard calculus question, but is actually a beautiful dance of symmetry and periodicity.

We are given a continuous function $f(x)$ that satisfies the functional equation $f(x)+f(x+1)=2$. Our mission is to find the value of $I_1+2I_2$, where $I_1={\int }_0^{8}f(x)dx$ and $I_2={\int }_{-1}^{3}f(x)dx$.

At first glance, this equation $f(x)+f(x+1)=2$ seems restrictive, but it is actually a gift. It tells us exactly how the function behaves as we move along the $x$-axis.

To see the pattern, let's perform a little algebraic manipulation. If we replace $x$ with $x+1$ in our original equation, we get:

Now, we have two equations. If we subtract the original equation, $f(x)+f(x+1)=2$, from this new one, the $f(x+1)$ terms vanish into thin air:

This is the breakthrough! The function $f(x)$ is periodic with a fundamental period $T=2$. Geometrically, this means the graph of the function repeats its shape every $2$ units. This is the heartbeat of the problem.

Now that we know the function repeats every $2$ units, we need to understand the area under the curve for one full period. Let's look at the integral from $0$ to $2$:

For the second integral, let's use a substitution. Let $x=t+1$, which implies $dx=dt$. When $x=1$, $t=0$, and when $x=2$, $t=1$.

The integral becomes:

Since $t$ is just a dummy variable, we can write this as ${\int }_0^{1}f(x+1)dx$. Now, combine the two parts back together:

Look at the integrand! We know $f(x)+f(x+1)=2$. So, the integral simplifies beautifully:

We have found the "base value" of the area for any interval of length $2$. It is exactly $2$.

Now, we calculate $I_1$ and $I_2$. $I_1={\int }_0^{8}f(x)dx$. The interval length is $8$, which is $4$ periods of length $2$.

Thus, $I_1=4\times 2=8$.

Similarly, $I_2={\int }_{-1}^{3}f(x)dx$. The interval length is $3-(-1)=4$, which is $2$ periods of length $2$.

Thus, $I_2=2\times 2=4$.

Finally, we compute the expression $I_1+2I_2$:

And there we have it! By identifying the hidden periodicity, we turned a complex integration problem into a simple arithmetic exercise. Keep this logic in your toolkit—whenever you see functional equations, look for the symmetry! The final answer is 16.