Key Concepts and Formulas

• Leibniz Integral Rule: If $G(t) = \int_{a(t)}^{b(t)} g(x) \, dx$, then $G'(t) = g(b(t)) \cdot b'(t) - g(a(t)) \cdot a'(t)$, provided $g(x)$ is continuous and $a(t), b(t)$ are differentiable.
• Differentiation of Powers: $\frac{d}{dt}(t^n) = nt^{n-1}$.
• Substitution: If $y = x^2$, then $x = \sqrt{y}$ for $x>0$.

Step-by-Step Solution

Step 1: Apply the Leibniz Integral Rule to Differentiate the Given Equation

We are given the equation: $\int\limits_0^{t^{2}}\left(f(x)+x^{2}\right) d x=\frac{4}{3} t^{3}, \quad \forall t > 0$ The purpose of this step is to eliminate the integral and obtain an equation involving $f(t^2)$ and $t$. We will differentiate both sides of the equation with respect to $t$.

The left-hand side (LHS) is an integral of the form $\int_{a(t)}^{b(t)} g(x) \, dx$, where:

• $g(x) = f(x) + x^2$
• $b(t) = t^2$ (upper limit)
• $a(t) = 0$ (lower limit)

Using the Leibniz Integral Rule, the derivative of the LHS with respect to $t$ is: $\frac{d}{dt} \left[ \int\limits_0^{t^{2}}\left(f(x)+x^{2}\right) d x \right] = g(b(t)) \cdot b'(t) - g(a(t)) \cdot a'(t)$ First, find the derivatives of the limits:

• $b'(t) = \frac{d}{dt}(t^2) = 2t$
• $a'(t) = \frac{d}{dt}(0) = 0$

Next, substitute the limits into the integrand $g(x) = f(x) + x^2$:

• $g(b(t)) = g(t^2) = f(t^2) + (t^2)^2 = f(t^2) + t^4$
• $g(a(t)) = g(0) = f(0) + (0)^2 = f(0)$

Now, apply the Leibniz Rule formula: $\frac{d}{dt} \left[ \text{LHS} \right] = (f(t^2) + t^4) \cdot (2t) - (f(0)) \cdot (0) = 2t(f(t^2) + t^4)$ The right-hand side (RHS) is $\frac{4}{3} t^3$. Its derivative with respect to $t$ is: $\frac{d}{dt} \left( \frac{4}{3} t^3 \right) = \frac{4}{3} \cdot 3t^2 = 4t^2$ Equating the derivatives of the LHS and RHS: $2t(f(t^2) + t^4) = 4t^2$

Step 2: Solve for $f(t^2)$

We have the equation: $2t(f(t^2) + t^4) = 4t^2$ Since the problem states $t > 0$, we can divide both sides by $2t$: $f(t^2) + t^4 = \frac{4t^2}{2t}$ $f(t^2) + t^4 = 2t$ Now, isolate $f(t^2)$: $f(t^2) = 2t - t^4$ This equation expresses the value of the function $f$ when its argument is $t^2$.

Step 3: Evaluate $f\left(\frac{\pi^{2}}{4}\right)$

We need to find the value of $f$ at the specific point $\frac{\pi^2}{4}$. From the equation $f(t^2) = 2t - t^4$, we need to find the value of $t$ such that $t^2 = \frac{\pi^2}{4}$. $t^2 = \frac{\pi^2}{4}$ Since $t > 0$, we take the positive square root: $t = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2}$ Now, substitute $t = \frac{\pi}{2}$ into the expression for $f(t^2)$: $f\left(\left(\frac{\pi}{2}\right)^2\right) = 2\left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^4$ $f\left(\frac{\pi^2}{4}\right) = \pi - \frac{\pi^4}{16}$ To match the format of the options, we can factor out $\pi$: $f\left(\frac{\pi^2}{4}\right) = \pi \left(1 - \frac{\pi^3}{16}\right)$

Common Mistakes & Tips

• Chain Rule Application: Ensure the derivative of the upper limit ($2t$) is correctly multiplied with the integrand evaluated at the upper limit. Forgetting this is a common error.
• Substitution Errors: When solving for $t$, remember that $t>0$ is given, so only the positive root is considered.
• Algebraic Manipulation: Pay close attention to signs and exponents when simplifying the equation after differentiation.

Summary

The problem was solved by applying the Leibniz Integral Rule to differentiate the given integral equation, which allowed us to obtain an explicit expression for $f(t^2)$. By setting $t^2$ equal to the desired argument $\frac{\pi^2}{4}$, we found the corresponding value of $t$ and substituted it into the derived expression for $f(t^2)$ to find $f\left(\frac{\pi^{2}}{4}\right)$.

The final answer is $\boxed{\pi\left(1-\frac{\pi^{3}}{16}\right)}$, which corresponds to option (B).