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

We are given a continuous function \( f \) satisfying the equation:

\(\int_{0}^{t^2} (f(x) + x^2) \, dx = \frac{4}{3} t^3 \) for all \( t > 0.\) 

We need to find the value of \( f\left(\frac{\pi^2}{4}\right) \).

Step-by-Step Solution:

1. Start by differentiating both sides of the equation with respect to \( t \) using the Leibniz rule for differentiation under the integral sign. The Leibniz rule states:

\(\frac{d}{dt} \int_{a(t)}^{b(t)} g(x, t) \, dx = g(b(t), t) \cdot b'(t) - g(a(t), t) \cdot a'(t) + \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} g(x, t) \, dx\)

1. For our specific integral:

\(\int_{0}^{t^2}(f(x) + x^2) \, dx\\) we have \( a(t) = 0 \) and \( b(t) = t^2 \). Therefore:

\(\frac{d}{dt} \int_{0}^{t^2}(f(x) + x^2) \, dx = (f(t^2) + (t^2)^2) \cdot 2t\)

1. Differentiate the right-hand side of the equation:

\(\frac{d}{dt} \left(\frac{4}{3} t^3 \right) = 4t^2\)

1. Equate the derivatives:

\((f(t^2) + t^4) \cdot 2t = 4t^2\)

\(2t f(t^2) + 2t^5 = 4t^2\)

1. Simplify and solve for \( f(t^2) \):

\(2t f(t^2) = 4t^2 - 2t^5\)

\(f(t^2) = 2t - t^4\)

1. Substitute \( t^2 = \frac{\pi^2}{4} \), which implies \( t = \frac{\pi}{2} \).
2. Substitute \( t = \frac{\pi}{2} \) into \( f(t^2) = 2t - t^4 \):

\(f\left(\frac{\pi^2}{4}\right) = 2 \left(\frac{\pi}{2}\right) - \left(\frac{\pi}{2}\right)^4\)

\(= \pi - \frac{\pi^4}{16}\)

Conclusion:

The value of \( f\left(\frac{\pi^2}{4}\right) \) is \(\pi \left(1 - \frac{\pi^3}{16}\right)\), which matches the given correct option.