🎥 Video solution / Text Solution of this question is given below:
$6\int_{1}^{x} f(t),dt = 3x f(x) + x^3 - 4$
Differentiate both side
$6f(x) = 3x f'(x) + 3f(x) + 3x^2$
$\Rightarrow 3f(x) = 3x f'(x) + 3x^2$
$\Rightarrow x f'(x) - f(x) = -x^2$
$\Rightarrow x\frac{dy}{dx} - y = -x^2$
$\Rightarrow \frac{d}{dx}\left(\frac{y}{x}\right) = -1$
$\Rightarrow \frac{y}{x} = -x + C$
$\Rightarrow f(x) = -x^2 + Cx$
At $x = 1,; y = 1 \Rightarrow C = 2$
$\Rightarrow f(x) = -x^2 + 2x$
$f(2) - f(3) = ( -4 + 4 ) - ( -9 + 6 ) = 3$