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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?

  • 0
    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$2012-12-06

1 Answers 1

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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.