Find derivative of integrate square function

Question

I am finding a solution of that function. Could you have me to resolve it

$$F=\left( \int {(ax+b-c)}^2 dx \right) +\lambda_1(a-m)^2+\lambda_2(b-n)^2$$

where $c,m,n ,\lambda_1,\lambda_2$ are constant

How to find

$$ \frac{\partial F}{\partial a}=?$$

$$ \frac{\partial F}{\partial b}=?$$ $$ \frac{\partial F}{\partial x}=?$$

Update: Sorry, I just one more require

How to find $a,b,x$ if I set

$$ \frac{\partial F}{\partial a}=0$$

$$ \frac{\partial F}{\partial b}=0$$ $$ \frac{\partial F}{\partial x}=0$$

Answer 1

Hint. You may first expand the integral.

Alternatively, you may observe that $$ \begin{align} \partial_a\left( \int {(ax+b-c)}^2 \:dx\right)& = 2\int x(ax+b-c)\:dx\\ \partial_b\left( \int {(ax+b-c)}^2 \:dx\right)&= 2\int (ax+b-c)\:dx\\ \partial_x\left( \int {(ax+b-c)}^2 \:dx\right)&= 2\int a(ax+b-c)\:dx. \end{align} $$ Then expand.

Answer 2

Regarding the first problem $\frac{\partial F}{\partial a}$. I am sure you can compute the other two, since they are very similar. If the integral boundaries are not depening on $a$ you can interchange integration and differentiation. Furthermore, assuming we are integrating with respect to $x$ we find $$\frac{\partial F}{\partial a}=2\int x(ax+b-c)dx + 2\lambda_1(a-m) $$