# FTC saves the day!

Problem-1 Given

$f(x) = \int\limits_{0}^{2x}f(\frac{t}{2})\;dt +\log(2)\quad\quad\quad(\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

Let

$p=2x \implies \frac{dp}{dx} = 2.\quad\quad\quad(1-1)$

By $(\star)$,

$\frac{df(x)}{dx} =\frac{d}{dx} \int\limits_{0}^{p} f(\frac{t}{2})\;dt + \frac{d\log(2)}{dx} = \underline{\frac{d}{dp}\left(\int\limits_{0}^{p} f(\frac{t}{2})\;dt\right)} \cdot \frac{dp}{dx}\overset{\textbf{FTC}}{=}\underline{f(\frac{p}{2})}\cdot \frac{dp}{dx}\overset{(1-1)}{=}2f(x),$

i.e.,

$\frac{df(x)}{dx} = 2f(x).\quad\quad\quad(1-2)$

Moreover, we see from $(\star)$ that

$f(0) = \int\limits_{0}^{0}f(\frac{t}{2})\;dt + \log(2) = 0 + \log(2) = \log(2).\quad\quad\quad(1-3)$

Solving initial-value problem

$\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}$

gives

$f(x) = \log(2)\cdot e^{2x}.$

We use Omega CAS Explorer to verify:

Fig. 1-1

Problem-2 Given

$\int\limits_{0}^{1}f(u\cdot x) \;du = \frac{1}{2} f(x) +1\quad\quad\quad(\star\star)$

where $f(x)$ is a continuous function, find $f(x).$

Solution

Let $p=u\cdot x,$

$u=\frac{p}{x} \implies \frac{du}{dp} = \frac{1}{x}\quad\quad\quad(2-1)$

$u=0\implies p=0; u=1\implies p=x.\quad\quad\quad(2-2)$

$\int\limits_{0}^{1}f(u\cdot x)\;du\overset{(2)}{=} \int\limits_{0}^{x}f(p)\cdot\frac{du}{dp}\cdot dp\overset{(1)}{=}\int\limits_{0}^{x}f(p)\frac{1}{x}\;dp=\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp.\quad\quad\quad(2-3)$

By (2-3), we express $(\star\star)$ as

$\frac{1}{x}\int\limits_{0}^{x}f(p)\;dp = \frac{1}{2}f(x)+1,$

i.e.,

$\int\limits_{0}^{x} f(p)\;dp = \frac{x}{2}f(x)+x.$

It follows that

$\underline{\frac{d}{dx}\left(\int\limits_{0}^{x}f(p)\;dp\right)}=\frac{d}{dx}\left(\frac{x}{2}f(x)+x\right)\overset{\textbf{FTC}}{\implies}\underline{f(x)}=\frac{1}{2}\left(f(x) + x\frac{d f(x)}{dx}\right)+1.\;(2-4)$

Solving differential equation (2-4) (see Fig. 2-1) gives

$f(x) = c x + 2.$

Fig. 2-1

The solution is verified by Omega CAS Explorer:

Fig. 2-2

Exercise-1 Solving $\begin{cases} \frac{df(x)}{dx} = 2f(x)\\ f(0)=\log(2)\end{cases}$ using a CAS.

Exercise-2 Solving (2-4) without using a CAS.