Fundamental Theorem of Calculus when Integrand is a Function of the Bounds I have a function, 
  $$F(r) = \int_0^r |c x^2 + {(2 a + b - 4 a r - 3 b r - 2 c r) x^2\over2 r} + b x^3 +
  a x^4| dx$$
a, b and c are constants. I want to determine r such that $f=F'(r) = k$. Integrating with an absolute value is nasty, so my first thought was to use the First Fundamental Theorem of Calculus, which states:
if $F(x) = \int_a^x f(t) dt$ then $F'(x) = f(x)$
But what I have is more like $F(x) = \int_a^x f(t, x)dt$ and so I'm not quite sure if/how the theorem applies. Is there any way to compute $F'(r)$ without first solving for $F(r)$ (which requires breaking the integrand apart into a piecewise function to remove the absolute value)?

You can consider the function $G(x,y)=\int_a^y f(x,t)\ dt$. 
Then $\frac{\partial G}{\partial x}(x,y)=\int_a^y\frac{\partial f}{\partial x}(x,t)\ dt$ and $\frac{\partial G}{\partial y}(x,y)=f(x,y)$.
And $F(r)=G(r,r)$ so $F'(r)=\frac{\partial G}{\partial x}(r,r)+\frac{\partial G}{\partial x}(r,r)=\int_a^r\frac{\partial f}{\partial x}(r,t)\ dt+f(r,r)$ by the Chain Rule.
This works on intervals where $\frac{\partial f}{\partial x}(r,t)$ exists and is continuous.
