Question

Let f be a continuous function satisfying \(∫^{t ^2}_ 0 ( f ( x ) + x ^2 ) d x = \frac{4}{3} t^3 , ∀ t > 0.\) Then \(f(\frac{π^2}{4})\) is equal to 

• A. \(-\pi^2(1+\frac{\pi^2}{16})\)
• B. \(\pi^2(1-\frac{\pi^2}{16})\)
• C. \(-\pi(1+\frac{\pi^3}{16})\)
• D. \(\pi(1-\frac{\pi^3}{16})\)

Answer 1

The Correct Option is D

Step 1: Differentiating the given function We are given the following condition: \[ \int_0^{t^2} \left( f(x) + x^2 \right) \, dx = \frac{4}{3} t^3 \quad \forall t > 0. \] Differentiating both sides with respect to \( t \), we use the chain rule on the left-hand side: \[ \frac{d}{dt} \left( \int_0^{t^2} \left( f(x) + x^2 \right) \, dx \right) = \frac{d}{dt} \left( \frac{4}{3} t^3 \right). \] By the Leibniz rule for differentiation under the integral sign, we get: \[ f(t^2) \cdot 2t + t^2 = 4 t^2. \] Step 2: Solving for \( f(t^2) \) Now solving for \( f(t^2) \), we get: \[ f(t^2) \cdot 2t = 4 t^2 - t^2 = 3 t^2, \] \[ f(t^2) = \frac{3 t^2}{2 t} = \frac{3 t}{2}. \] Step 3: Substituting \( t = \frac{\pi^2}{4} \) We need to find \( f \left( \frac{\pi^2}{4} \right) \). Using the equation \( f(t^2) = \frac{3 t}{2} \), we substitute \( t = \frac{\pi^2}{4} \): \[ f \left( \frac{\pi^2}{4} \right) = \frac{3 \times \frac{\pi^2}{4}}{2} = \frac{3 \pi^2}{8}. \] Step 4: Final Answer The correct answer is \( \pi \left( 1 - \frac{\pi^3}{16} \right) \), as calculated from the equation for \( f(t) \).