Derivative of an accumulation function with variable lower limit

Since $x$ appears as the lower limit, first rewrite the integral using the property $\int_x^5 f(t)\,dt = -\int_5^x f(t)\,dt$. So $F(x) = -\int_5^x (t^2+4)\,dt.$

Now apply FTC Part 1 to $\int_5^x (t^2+4)\,dt$: its derivative with respect to $x$ is simply $x^2+4$ (the integrand evaluated at the upper limit, since the upper limit is just $x$ itself). Therefore $F'(x) = -(x^2+4) = -x^2 - 4.$

The answer is negative. Swapping the limits of an integral, using the property that the integral from x to five equals the negative of the integral from five to x, introduces a minus sign in front of the expression. This negative sign must be carried through when applying FTC Part 1 to differentiate the rewritten accumulation function.