If \(6\int_{1}^{x} f(t)dt = 3x f(x) + x^3 - 4, x \geq 1\) then value of (f(2)-f(3)) is :
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Equations involving integrals can often be solved by differentiation using Leibniz's rule. Remember to find an initial condition from the original integral equation to determine the constant of integration.
Concept:
Moment of Inertia of a disc is \(I = \frac{1}{2}MR^2\).
Mass can be expressed as Density \(\times\) Volume: \(M = \rho (\pi R^2 t)\).
Substituting this:
\[ I = \frac{1}{2} (\rho \pi R^2 t) R^2 \propto t R^4 \]
Step 1: Equate the moments of inertia.
\[ t_1 R_1^4 = t_2 R_2^4 \]
Step 2: Solve for ratio of thicknesses.
\[ \frac{t_1}{t_2} = \left( \frac{R_2}{R_1} \right)^4 \]
Given \(\frac{R_1}{R_2} = \frac{1}{2}\), so \(\frac{R_2}{R_1} = 2\).
\[ \frac{t_1}{t_2} = (2)^4 = 16 \]
\[
\boxed{\dfrac{t_1}{t_2} = 16}
\]
