Calculate the Derivative of a univariable integral at a point $4$

Question

Considering the function below:

the objective is to calculate $F'(4)$ (the derivative of $F(x)$ in the point $4$)?

we know that:

and that:

so if I try to replace $x$ by $4$ in $F'(x)$ I get $0$. Is that correct?

Answer 1

Differentiate using the product rule and the Fundamental Theorem of Caclulus

(see 'Part 1' here: http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Formal_statements)

gives:

$$F'(x) = -\frac{2}{x^3} \int_4^x \dots dx + \frac{1}{x^2}(4x^2-2F'(x)).$$

Plug in the value $x=4$ and the integral disappears (as you say in your question); we are left with

$$F'(4) = \frac{1}{16} (4\cdot 16 - 2F'(4)).$$

Solving this equation gives $F'(4) = \frac{32}{9}.$