Question

Derive Integral equation for dy/dv

Answer 1

By the fundamental theorem of calculus,

$\displaystyle y = \int_0^v \sqrt{3+4t^2} \, dt \implies \boxed{\frac{dy}{dv} = \sqrt{3+4v^2}}$

In general,

$\displaystyle \frac{d}{dx} \int_0^{g(x)} f(u) \, du = f(g(x)) \frac{dg}{dx}$