Question of the week #4: Let a continuous real function $f$ and $f(x)=e^{\int_{0}^{x}f(t)dt}$ for all $x<1$. Find the formula of the function $f$.
Solution: First we note that $f(0)=e^{\int_{0}^{0}f(t)dt}=e^{0}=1$. Moreover, we can see that
$$
f(x)>0, \ \ \forall x<1
$$
For any $x<1$ we have:
$$
\begin{array}{c}
f^{\prime}(x)=\bigg( e^{\int_{0}^{x}f(t)dt} \bigg) ^{\prime}=e^{\int_{0}^{x}f(t)dt} \big( \int_{0}^{x}f(t)dt \big)^{\prime}=f(x) \cdot f(x) \Leftrightarrow \\
\\
\Leftrightarrow f^{\prime}(x)=f^{2}(x) \Leftrightarrow \frac{f^{\prime}(x)}{f^{2}(x)} =1 \Leftrightarrow \\
\\
\Leftrightarrow \bigg( -\frac{1}{f(x)} \bigg)^{\prime}=(x)^{\prime} \Leftrightarrow -\frac{1}{f(x)}=x+c
\end{array}
$$
where $c \in \mathbb{R}$ is the constant of integration.
But the above relation readily implies (for $x=0$) that $c=-\frac{1}{f(0)}=-1$.
So we finally get:
$$
f(x)=\frac{1}{1-x}, \ \ \forall x<1
$$
Solution: First we note that $f(0)=e^{\int_{0}^{0}f(t)dt}=e^{0}=1$. Moreover, we can see that
$$
f(x)>0, \ \ \forall x<1
$$
For any $x<1$ we have:
$$
\begin{array}{c}
f^{\prime}(x)=\bigg( e^{\int_{0}^{x}f(t)dt} \bigg) ^{\prime}=e^{\int_{0}^{x}f(t)dt} \big( \int_{0}^{x}f(t)dt \big)^{\prime}=f(x) \cdot f(x) \Leftrightarrow \\
\\
\Leftrightarrow f^{\prime}(x)=f^{2}(x) \Leftrightarrow \frac{f^{\prime}(x)}{f^{2}(x)} =1 \Leftrightarrow \\
\\
\Leftrightarrow \bigg( -\frac{1}{f(x)} \bigg)^{\prime}=(x)^{\prime} \Leftrightarrow -\frac{1}{f(x)}=x+c
\end{array}
$$
where $c \in \mathbb{R}$ is the constant of integration.
But the above relation readily implies (for $x=0$) that $c=-\frac{1}{f(0)}=-1$.
So we finally get:
$$
f(x)=\frac{1}{1-x}, \ \ \forall x<1
$$
