Animated Solution for Mathematics - Definite Integration: Let $f(x)$ be a positive function and $I_1={\int }_{1/2}^{1}2xf(2x(1-2x))dx$ and $I_2={\int }_{-1}^{2}f(x(1-x))dx$. Then the value of $\frac{I_2}{I_1}$ is equal to ________

Welcome, fellow explorer of the mathematical landscape. Today, we are going to tackle a problem that might look like a tangled mess of integrals at first glance, but beneath the surface, it hides a beautiful, elegant structure.

We are given two integrals:

We are tasked with finding their ratio. The key to this problem is not brute force, but rather, the art of observation.

Simplifying the Complexity

Let us begin by looking at $I_1$. The argument $2x(1-2x)$ is quite complex.

Let us use the substitution $2x=t$. This implies $2dx=dt$, or $dx=\frac{dt}{2}$.

When $x=1/2$, $t=1$. When $x=1$, $t=2$. Substituting these into our integral, we get:

To make it even more intuitive, let us define a new function $g(x)=f(x(1-x))$. Now, $I_1$ simplifies to:

Unveiling the Symmetry

Now, let us turn our attention to $I_2={\int }_{-1}^{2}f(x(1-x))dx$. Using our new definition, this is simply:

Let us test the symmetry of $g(x)$ by evaluating $g(1-x)$:

This tells us that $g(x)$ is perfectly symmetric about the line $x=1/2$. This symmetry is the heartbeat of the problem.

The King's Property

We have $I_1$ involving ${\int }_1^{2}xg(x)dx$ and $I_2$ involving ${\int }_{-1}^{2}g(x)dx$. To bridge this gap, let us define a helper integral $J={\int }_{-1}^{2}xg(x)dx$.

We invoke the King's Property: ${\int }_a^{b}h(x)dx={\int }_a^{b}h(a+b-x)dx$. Here, $a+b=-1+2=1$.

So, $J={\int }_{-1}^2(1-x)g(1-x)dx$. Because $g(1-x)=g(x)$, we have:

Notice that the first term is $I_2$ and the second term is $J$ itself. Thus, $J=I_2-J$, which simplifies to $2J=I_2$, or $J=\frac{1}{2}{I}_2$.

The Final Connection

We have $I_2=2J=2{\int }_{-1}^{2}xg(x)dx$. Due to the symmetry of $g(x)$ around $x=1/2$, the integral ${\int }_{-1}^{2}xg(x)dx$ is equal to ${\int }_1^{2}xg(x)dx$.

Substituting this back, we get:

From our earlier work, we know $I_1=\frac{1}{2}{\int }_1^{2}xg(x)dx$, which implies ${\int }_1^{2}xg(x)dx=2I_1$.

Finally, substituting this into our expression for $I_2$:

Therefore, the ratio is:

The logic holds, and the final result is 4.