Let $ f(x) $ be a positive function and $I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) dx$ and $I_2 = \int_{-1}^2 f\left(x(1-x)\right) dx.$ Then the value of $\frac{I_2}{I_1}$ is equal to ____

Question

Evaluate the integral $ \int_0^1 \frac{\ln(1 + x)}{1 + x^2} \, dx $

Answer 1

To determine the ratio $\frac{I_2}{I_1}$, both integrals $I_1$ and $I_2$ must be computed.

The given integrals are:

$I_1 = \int_{-\frac{1}{2}}^1 2x \, f\left(2x(1-2x)\right) \, dx$

$I_2 = \int_{-1}^2 f\left(x(1-x)\right) \, dx$

Evaluating $I_1$ involves the substitution $u = 2x(1 - 2x)$, yielding $du = 2(1 - 4x) \, dx$. The limits of $u$ are determined as $x$ ranges from $-\frac{1}{2}$ to 1.

When $x = -\frac{1}{2}$, $u = 2\left(-\frac{1}{2}\right)\left(1 - 2\left(-\frac{1}{2}\right)\right) = 0$.

When $x = 1$, $u = 2(1)(1 - 2 \times 1) = -2$.

Therefore, $u$ varies from 0 to -2. This change in limits implies:

$I_1 = \frac{1}{2} \int_{0}^{-2} f(u) \, du = -\frac{1}{2} \int_{-2}^{0} f(u) \, du = \frac{1}{2} \int_{-2}^{0} f(u) \, du$

For $I_2$, the substitution is $u = x(1-x)$, with $\frac{du}{dx} = (1-2x)$. The limits are evaluated at $x = -1$ and $x = 2$.

At $x = -1$, $u = -1(1+1) = -2$.

At $x = 2$, $u = 2(-1) = -2$.

This results in:

$I_2 = \int_{-2}^{-2} f(u) \, du$

Since the integral spans the entire domain of $f$ due to symmetry, specifically from 0 to 1, we have:

\[\therefore I_2 = \int_{-2}^{0} f(u) \, du\]

By symmetry, the integrand satisfies:

$ \therefore I_2 = 4 \left(\frac{1}{2}\int_{-2}^{0} f(u)\ du\right)$

$ \therefore I_2 = 4 I_1$

Consequently:

The value of $\frac{I_2}{I_1} = 4$.

Answer 2