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So, I am preparing myself for the math subject GRE with a practice test. Not only did I bomb the test, but I found that even things I thought I understood now seem confusing.

Case in point: the first question, the question that ought to be the easiest. It asked me to compute the derivative with respect to x of:

$$\int_e^x \log t \;dt$$

I can't post images yet so please click here

So, I thought using the Fundamental Theorem of Calculus, I'd get log(x) - log(e) = log (x) - 1. That wasn't even a choice.

Not only that, but the solution guide thought the answer was TOO SIMPLE TO EVEN EXPLAIN other than saying it's the FTC. The answer is log(x).

Please explain where my reasoning is flawed, I would be very grateful.

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  • $\begingroup$ My apologies, I edited the question $\endgroup$ – madhatter5 Aug 17 '17 at 23:35
  • $\begingroup$ Well, that is the FTC. How would you state that theorem? $\endgroup$ – lulu Aug 17 '17 at 23:36
  • $\begingroup$ Keep in mind that changing the lower limit, say to $2$, only changes the function by an additive constant (namely the area under the graph between $2$ and $e$). $\endgroup$ – lulu Aug 17 '17 at 23:38
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    $\begingroup$ @madhatter5 The derivative with respect to $x$ of $F(b)-F(a)$ is $0$. The derivative with respect to $x$ of $F(x)-F(a)$ is $f(x)$. $\endgroup$ – John Griffin Aug 17 '17 at 23:45
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    $\begingroup$ Remember, $F(x)$ is a function; $F(a)$ is a constant. $\endgroup$ – G Tony Jacobs Aug 17 '17 at 23:50
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Let $F(x)$ be an antiderivative of $\log(x)$. Then the integral evaluates to $F(x)-F(e)$. Taking the derivative of that, you get $\log(x)-0$, because the derivative of a constant is zero.

Now, applying the FTC even more directly, the formula is this:

$$\frac{d}{dx}\int_a^x f(t) \,\,dt = f(x),$$

without an $f(a)$ subtracted.

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By the fundamental theorem of calculus, if

$$F(x) = \int_{a}^x f(u)\,du$$

(where the upper limit of the integration is the only part of the expression that depends on $F$'s argument)

then $$\frac{dF}{dx} = f(x).$$

In this case, the result would be $\frac{dF}{dx} = \log{(x)}$.


If you want to see this in a more computational way, suppose you have

$$F(x) = \int_a^x f(u) \,du$$

and let $g(u)$ be an antiderivative of $f(u)$ (so that $g(u) =\int f(u)\,du$ and conversely $g^\prime = f$). We should have that

$$\begin{align*} F(x) &= \int_a^x f(u) \, du\\ F(x) &= g(x) - g(a)\\ \frac{\partial F}{\partial x} &= \frac{d}{dx}\left[g(x) - g(a)\right]\\ \frac{\partial F}{\partial x} &= \frac{dg}{dx} - 0\\ \frac{\partial F}{\partial x} &= f(x)\\ \end{align*}$$

That is, I think your only mistake was using $f$ in a place where you meant $g$, the antiderivative of $f$.

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Using the Fundamental Theorem of Calculus you are supposed to get a continuous function $F(x)$ such that $F(x)-F(e) = \int_e^x \log(t)dt$. As $\log$ is continuous, it follows that $\dfrac{dF}{dt} = log(t)$. So by differentiating the first equation you get $log(t)$.

Hope it helps!

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  • $\begingroup$ Thanks all! I understand now. $\endgroup$ – madhatter5 Aug 18 '17 at 14:09
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It is precisely FTC that answers your question, as stated in the previous answers. ..

A classic example of this is the definition of the natural logarithm :$$ \ln(x)=\int_1^x\frac1tdt $$. By FTC we now have an antiderivative for $\frac1x $...

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