I have come across this as a fundamental theorem of calculus:

$\frac{d}{dx}\int_{a}^{x} f(t) dt = f(x)\tag{1}$

for any constant $a$.

For example here:


I find myself wanting a solution to this however:

$\frac{d}{dx}\int_{-x}^{x} f(t) dt$

and am not finding this situation covered anywhere alas. I could of course easily rewrite this as:

$\frac{d}{dx}\int_{-x}^{0} f(t) dt + \frac{d}{dx}\int_{0}^{x} f(t) dt$

and this oft encountered fundamental theorem of calculus reduces this to:

$\frac{d}{dx}\int_{-x}^{0} f(t) dt + f(x)$

And if I knew $f(x)$ was symmetrical about 0, of course this would easily reduce to $2f(x)$ but I don't know that, in fact I know it's not. What then? Are there any further rules or theorems of calculus that I've forgotten that might help here?


Now I'm truly bamboozled and pulling my hair out. Somewhere I am making a fundamentally simple error it seems but I cannot after review, review, and review find where. To wit here is the problem:

From basic integral conventions we know:

$\begin{align} \int_a^c f(x) \, dx &{}= \int_a^b f(x) \, dx - \int_c^b f(x) \, dx \tag{2}\\ &{} = \int_a^b f(x) \, dx + \int_b^c f(x) \, dx\tag{3} \end{align}$


$\int_a^b f(x) \, dx = - \int_b^a f(x) \, dx\tag{4}$

Applying these we can write:

$\begin{align*} \frac{d}{dx}\int_{-x}^{x} f(t)\;dt &= \frac{d}{dx}\int_{-x}^{0} f(t)\;dt + \frac{d}{dx}\int_{0}^{x} f(t)\;dt \\ &= -\frac{d}{dx}\int_{0}^{-x} f(t)\;dt + \frac{d}{dx}\int_{0}^{x} f(t)\;dt \\ &= \frac{d}{dx}\int_{0}^{x} f(t)\;dt - \frac{d}{dx}\int_{0}^{-x} f(t)\;dt \\ &= f(x)- f(-x) \end{align*}$

But, and here's where I pull my hair out looking over and over for a sign error above or below, from the Leibniz integral rule, we can write:

$f_1(x) = -x \Rightarrow f^\prime_1(x) = -1$

$f_2(x) = x \Rightarrow f^\prime_2(x) = 1$

And so:

$\frac{d}{dx} \left(\int_{-x}^{x} g(t) \,dt \right ) = g(x) + g(-x)$

Two conflicting results! Where oh where have I gone wrong? I'm sure I'll be kicking myself soon if you help me spot it - which will be a step up from pulling my hair out.

  • 2
    $\begingroup$ For a generalization see, e.g., Leibniz integral rule. $\endgroup$ – user539887 Jun 20 '18 at 10:58
  • $\begingroup$ Updated with a followup, as I tried that and some standard conventions too and get conflicting results ... $\endgroup$ – Bernd Wechner Jun 21 '18 at 4:00
  • $\begingroup$ Between the displayed lines 10 and 11: $-\frac{d}{dx}\int\limits_{0}^{-x}f(t)\,dt$ equals $-(-1)\cdot f(-x)=f(-x)$. $\endgroup$ – user539887 Jun 21 '18 at 7:03
  • $\begingroup$ Not seeing where your $(-1)$ comes from alas. The fundamental theorem is: $\frac{d}{dx}\int_{a}^{x} f(t) dt = f(x)$ to wit $\frac{d}{dx}\int_{a}^{-x} f(t) dt = f(-x)$. Whence cometh the $-1$? $\endgroup$ – Bernd Wechner Jun 21 '18 at 10:05

The fundamental theorem of calculus write $$\frac d {dx}\int_{a(x)}^{b(x)} f(t) \, dt=f(b(x))\, b'(x)-f(a(x))\, a'(x)$$

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If you write $F (x)=\int_a^xf (t)\,dt $, you are saying that $$\tag1( F(-x))'=F'(-x). $$ Which is wrong. For instance if $F (x)=x^2$, the equality $(1) $ becomes $2x=2 (-x) $.

When you write $F (-x) $, you have $F (h (x))$, with $h (x)=-x $. The derivative, obtained with the chain rule, is $$\tag2 (F (h (x))'=F'(h (x))\,h'(x). $$ That's where your missing $-1$ comes from.

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  • $\begingroup$ @user539887 resolves $-\frac{d}{dx}\int_{0}^{-x} f(t)\;dt$ differently, and thanks I think I can see why now, because equation 1 I abused by substituting -x for x in the bound, but failed to do so in the differential denominator. To wit I should have completed it as $\frac{d}{-dx}\int_{a}^{-x} f(t) dt = f(-x)$. And the dx is where that missing sign came from. The chain rule explains Leibniz's integral rule, but this explains where I went wrong in applying equation 1. Doh! I am no kicking myself. $\endgroup$ – Bernd Wechner Jun 21 '18 at 12:27

Let $F(x)$ be an antiderivative of $f(x)$. Then


Now by the chain rule,

$$\frac d{dx}\int_{-x}^xf(t)\,dt=(x)'f(x)-(-x)'f(-x)=f(x)+f(-x).$$

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user539887 found where the error was and forced a major rethink and focus on that one line, from which I finally realised, that I had abused equation 1 when I subsituted -x for x.

The complete substitution yields:

$\frac{d}{-dx}\int_{a}^{-x} f(t) dt = f(-x)$

which can be rearranged to:

$\frac{d}{dx}\int_{a}^{-x} f(t) dt = -f(-x)$

And so my second method should read:

$\begin{align*} \frac{d}{dx}\int_{-x}^{x} f(t)\;dt &= \frac{d}{dx}\int_{-x}^{0} f(t)\;dt + \frac{d}{dx}\int_{0}^{x} f(t)\;dt \\ &= -\frac{d}{dx}\int_{0}^{-x} f(t)\;dt + \frac{d}{dx}\int_{0}^{x} f(t)\;dt \\ &= \frac{d}{dx}\int_{0}^{x} f(t)\;dt + \frac{d}{dx}\int_{0}^{-x} f(t)\;dt \\ &= f(x) + f(-x) \end{align*}$

and it agrees with the Leibniz integration rule. All confusion abated.

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