# 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)?

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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.

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That is a very nice derivation. It's too bad it doesn't work with my particular function, since ∂f/∂x isn't continuous. –  Wallacoloo Apr 28 '14 at 6:59
Actually, I suppose I could find the discontinuities and integrate each portion individually to create a piecewise function. But that seems to be similar in complexity to the naive way of integrating this. –  Wallacoloo Apr 28 '14 at 7:08
Surely the only discontinuity is at $r=0$? Should be safe enough to assume that $r \neq 0$. If you have a function on, eg $[0,1]$, then when you differentiate, you can only claim that it is differential exists on $(0,1)$, since needs to be right and left differentiable. –  Smiley Sam Apr 28 '14 at 8:20
You can also use the standard $(F(r+h)-F(r))/h$ thing ($r0$ as above) and then just collect the terms internally and then use the known derivatives. :) –  Smiley Sam Apr 28 '14 at 8:23