Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
If you enclose the upper/lower limits with {}, Mathjax will render the properly, i.e. \int_{2x}^{3x} f(t) dt will produce $$\int_{2x}^{3x} f(t) dt$$ – Pragabhava Dec 6 '12 at 6:40
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.