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?

  • $\begingroup$ If you enclose the upper/lower limits with {}, 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, 2012 at 6:40

1 Answer 1


Let $$g(x)=\int_{\alpha(x)}^{\beta(x)} f(t) dt.$$ By FTC, $$ g(x) = F(\beta(x)) - F(\alpha(x)) $$ where $F$ is an anti-derivative of $f$. Then by chain rule \begin{align*}g'(x) &= (F(\beta(x)))' - (F(\alpha(x)))' \\ &= f(\beta(x))\beta'(x) - f(\alpha(x))\alpha'(x). \end{align*}

In your case, $\beta'$ and $\alpha'$ give you the factors $3$ and $2$ respectively.


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