Question of the week #4:

$$

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

$$

**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

$$

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