Number of Solutions to Definite Integral Equation in Interval

To find the total number of distinct values of $x \in [0, 1]$ satisfying the integral equation $\int_0^x \frac{t^2}{1+t^4} dt = 2x - 1$

we define a function $f:[0, 1] \to \mathbb{R}$ as follows: $f(x) = \int_0^x \frac{t^2}{1+t^4} dt - 2x + 1$

We need to find the number of real roots of $f(x) = 0$ in the interval $x \in [0, 1]$.

Step 1: Check the continuity and differentiability of $f(x)$ By the Fundamental Theorem of Calculus, $f(x)$ is continuous and differentiable on $[0, 1]$.

Step 2: Determine the signs of $f(x)$ at the endpoints

• At $x = 0$: $f(0) = \int_0^0 \frac{t^2}{1+t^4} dt - 2(0) + 1 = 0 - 0 + 1 = 1 > 0$

• At $x = 1$: $f(1) = \int_0^1 \frac{t^2}{1+t^4} dt - 2(1) + 1 = \int_0^1 \frac{t^2}{1+t^4} dt - 1$

  For $t \in [0, 1]$, we have $1 + t^4 > 1$, which implies: $\frac{t^2}{1+t^4} < t^2$ Integrating both sides from $0$ to $1$: $\int_0^1 \frac{t^2}{1+t^4} dt < \int_0^1 t^2 dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3}$

  Therefore: $f(1) < \frac{1}{3} - 1 = -\frac{2}{3} < 0$

Since $f(0) > 0$ and $f(1) < 0$, by the Intermediate Value Theorem, there exists at least one root $x \in (0, 1)$ such that $f(x) = 0$.

Step 3: Check the monotonicity of $f(x)$ Differentiating $f(x)$ with respect to $x$ using the Leibniz Rule: $f'(x) = \frac{x^2}{1+x^4} - 2$

By the AM-GM inequality, for any real $x$: $1 + x^4 \ge 2\sqrt{1 \cdot x^4} = 2x^2 \implies \frac{x^2}{1+x^4} \le \frac{1}{2}$

Thus, for all $x \in [0, 1]$: $f'(x) \le \frac{1}{2} - 2 = -\frac{3}{2} < 0$

Since $f'(x) < 0$ for all $x \in [0, 1]$, the function $f(x)$ is strictly decreasing on $[0, 1]$.

Conclusion: A continuous, strictly decreasing function can cross the x-axis at most once. Since $f(0) > 0$ and $f(1) < 0$, there exists exactly one value $x \in [0, 1]$ satisfying $f(x) = 0$.

Thus, the total number of distinct values is 1.