Let's say that $$\ g(x) = \int_a^b (u^2-1)/(u^2+1) du$$ where $a=2x$ and $b=3x $ (Sorry, I couldn't figure out how to properly MathJax this)
The question asks to find the derivative using FTC. I had two approaches to this problem, but one of them is missing a factor... let me explain:
My first approach:
$$ g'(x)=d/dx \left(\int_a^b (u^2-1)/(u^2+1)\right) = d/dx [ g(3x) - g(2x) ] = f(3x) - f(2x) $$
Plugging in arguments 3x and 2x in the integrand yielded $(9x^2-1) /(9x^2+1) - (4x^2-1)/(4x^2+1)$ . However, it is missing a factor of 3 in the first term and a factor of 2 in the second. I'm aware that these values come from the derivatives of the upper and lower limits respectively, but why didn't they show up?
My second approach was correct, however:
$$ \int_a^b = \int_0^b - \int_0^a$$ I'll just show the work for $\int_0^b$...
$ b=3x$ therefore $db/dx=3$
$$ g'(x)=d/dx \left(\int_a^b (u^2-1)/(u^2+1)\right) = g'(x)=d/db \left(\int_a^b (u^2-1)/(u^2+1)\right) db/dx = 3((u^2-1)/(u^2+1))$$
As you can see, in this approach, the factor of 3 shows up because there is $db/dx=3$
Where does approach one go wrong?