Parameterized Integration - Part 1

We have $\newcommand{\ddx}{\dfrac d{dx}} \begin{array} { r l } F(x) &= \displaystyle \int_0^x t^2 f'(t) \, dt - x^2 \int_0^x f'(t) \, dt \\ \phantom0 \\ \implies \dfrac{dF}{dx} &= \displaystyle \ddx \left( \int_0^x t^2 f'(t) \, dt \right) - \ddx \left( x^2 \int_0^x f'(t) \, dt \right) \end{array}$ Let us compute these two derivatives separately. $$ Using fundamental theorem of calculus (Part 1) , we get $\newcommand{\ddx}{\dfrac d{dx}} \ddx \left( \int_0^x t^2 f'(t) \, dt \right) = x^2 f'(x)$ And with FTC part 2 as well as chain rule , $\newcommand{\ddx} {\dfrac d{dx}} \ddx \left(x^2 \int_0^x f'(x) \, dt \right) = x^2 \cdot \underbrace{\ddx \left( \int_0^x f'(t) \, dt \right)}_{= f'(x)} + \underbrace{\left( \int_0^x f'(t) \, dt \right)}_{=\, f(x) - f(0) } \cdot \underbrace{\ddx (x^2) }_{=\, 2x} = x^2 f'(x) + 2x \, [ f(x) - f(0) \, ]$ Combining them together, we get $\dfrac{dF}{dx} = \cancelto{0}{ x^2 f'(x) - x^2 f'(x)} - 2x \, [ \, f(x) - f(0) \, ] = \boxed{ - 2x \, ( \, f(x) - f(0) \, ) }.$

$F(x) \; = \; -\int_0^x (x^2-t^2)f'(t)\,dt \; = \; -\int_0^x \left(\int_t^x 2s\,ds\right)\,f'(t)\,dt \; = \; -\int_0^x 2s\left(\int_0^s f'(t)\,dt\right)\,ds \: = \; -2\int_0^x s(f(s)-f(0))\,ds$ and hence $F'(x) = \boxed{-2x(f(x)-f(0))}$ .