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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?

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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

1 Answer 1

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.

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