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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:

http://mathmistakes.info/facts/CalculusFacts/learn/doi/doi.html

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?

UPDATE:

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:

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

and:

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

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.

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

$$\int_{-x}^xf(t)\,dt=\left.F(t)\right|_{t=-x}^x=F(x)-F(-x).$$

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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