Let's say that $$\ g(x) = \int_{2x}^{3x} \frac{(u^2-1)}{(u^2+1)} du$$
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:
$$ g'(x)=\frac{d}{dx} \left(\int_{2x}^{3x} \frac{(u^2-1)}{(u^2+1)}\right) = \frac{d}{dx} [ g(3x) - g(2x) ] = f(3x) - f(2x) $$
Plugging in arguments $3x$ and $2x$ in the integrand yielded $\frac{(9x^2-1)}{(9x^2+1)} - \frac{(4x^2-1)}{(4x^2+1)}$ . 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:
$$ \int_{2x}^{3x} = \int_0^{3x} - \int_0^{2x}$$ I'll just show the work for $\int_0^{3x}$...
therefore $\frac{d}{dx}3x=3$
$$ g'(x)=\frac{d}{dx} \left(\int_{2x}^{3x} \frac{(u^2-1)}{(u^2+1)}\right) = g'(x)=\frac{d}{db} \left(\int_{2x}^{3x} \frac{(u^2-1)}{(u^2+1)}\right) \frac{d}{dx}3x = 3\frac{(u^2-1)}{(u^2+1)}$$
As you can see, in this approach, the factor of 3 shows up because there is $\frac{d}{dx}3x=3$
Where does approach one go wrong?
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, Mathjax will render the properly, i.e.\int_{2x}^{3x} f(t) dt
will produce $$\int_{2x}^{3x} f(t) dt$$ $\endgroup$ – Pragabhava Dec 6 '12 at 6:40