# Integral of the derivative of a function

• May 1st 2013, 06:00 PM
SlipEternal
Integral of the derivative of a function
Suppose $f(x)$ is a differentiable function whose derivative is $\dfrac{1}{1+x^2}$. We know that $f(x) = \arctan{x}+C$. Now, suppose $g(t) = \tan(t^3+4t^2-2t+7)$ and $h(t) = \cot(t^3-7t^2-34t+1)$. Suppose we want to figure out what $f(h(t))-f(g(t))$ is. Taking the derivative, we get $f^\prime(h(t))h^\prime(t)-f^\prime(g(t))g^\prime(t) = 6(t+2)(t-3)$. Integrating, we get $2t^3-3t^2-36t+C$ which undoubtedly achieves values outside the range of $\arctan(x)$ regardless of the constants for each. Is this because $\tan(x)$ is discontinuous?
• May 1st 2013, 07:27 PM
chiro
Re: Integral of the derivative of a function
Hey SlipEternal.

The discontinuity in addition to the vertical asymptotes both contribute to this phenomena. You can't integrate something when you have these kinds of things involved.
• May 1st 2013, 08:36 PM
ibdutt
Re: Integral of the derivative of a function
You have f(x) = arc tan x + C
Now f(h(t)) = arc tan[ cot ( t^3 - 7t^2 - 34 t + 1 )] = arc tan[ tan { pi/2 - ( t^3 - 7t^2 - 34 t + 1 )}] = pi/2 - ( t^3 - 7t^2 - 34 t + 1 )
Similarly f(g(t)) = t^3 + 4t^2 - 2 t + 1
Now you can proceed to evaluate
f(h(t)) - f(g(t))
• May 2nd 2013, 04:46 AM
SlipEternal
Re: Integral of the derivative of a function
Quote:

Originally Posted by chiro
Hey SlipEternal.

The discontinuity in addition to the vertical asymptotes both contribute to this phenomena. You can't integrate something when you have these kinds of things involved.

So, if I let $A = \{t\in \mathbb{R}\mid n\in \mathbb{Z}, t^3+4t^2-2t+7 = (2n+1)\dfrac{\pi}{2} \}$, then I define $A^+ = \{t\in A \mid t\ge 0\}, A^- = \{t\in A \mid t < 0\}$, I can define an set of points $(a_n)_{n\in \mathbb{Z}}$ where $a_0 = \max A^-, a_{-n} = \max\left(A^- \setminus \{a_0,\ldots, a_{1-n}\}\right)$, $a_1 = \min A^+, a_n = \min\left(A^+ \setminus \{a_1,\ldots,a_{n-1}\}\right)$, I can integrate $f^\prime(g(t))g^\prime(t)$ over each interval $(a_{n-1},a_n)$, and in each case, I will get $t^3+4t^2-2t+C$. I can use a similar trick for $f^\prime(h(t))h^\prime(t)$. So, the integral works, and produces the correct result. But, my resulting function has a lot of points of discontinuity (the union of the points of discontinuity of $g(t)$ and $h(t)$ individually).

Now, $\sec^2 x$ has an infinite number of points of discontinuity, yet when I integrate that, I get $\tan x + C$ (here, there is a single constant that works for each interval). Yet, $f(x)$ and $f^\prime(x)$ both have no points of discontinuity, so the composition functions should only be discontinuous at values of $t$ that are points of discontinuity for $g(t)$ and/or $h(t)$, and each of those discontinuities for the composition functions "should" be removable discontinuities since $\displaystyle \lim_{x\to -\infty}f^\prime(x) = \lim_{x\to \infty} f^\prime(x) = 0$, which further confuses me as to why I get a piecewise continuous function where each piece may have a different constant of integration.

To better see my confusion, consider the function $j(t) = \tan(t)$. Now, $f^\prime(j(t))j^\prime(t) = 1$, and the integral would be $t+C$. So, it is the co-restriction of the domain that is confusing me. Would the actual formula be $f(j(t)) = t - \left\lfloor \dfrac{t+\frac{\pi}{2}}{\pi} \right\rfloor\pi + C$ where $\lfloor \cdot \rfloor$ is the floor function?