For $x \geq 1$ we define $f(x)$ as the unique number $c$ such that $c^c = x $. Evaluate the integral:
$$\mathcal{J}=\int_0^e f(x) \, {\rm d}x$$
Solution
Well for starters let me say that the special in the title does not stand for something WOW ! . It is just a category of exercises in computation of integrals. All we have to do is actually invoke the inverse function somehow.
Well,
\begin{align*}
\int_{0}^{e} f\left ( e^x \right ) \, {\rm d}x &\overset{e^x=y^y}{=\! =\! =\! =\!} \int_{1}^{e} y \left ( \ln y +1 \right ) \, {\rm d}y\\
&= \int_{1}^{e} y \ln y \, {\rm d}y + \int_{1}^{e} y \, {\rm d}y\\
&= \frac{1}{4} + \frac{e^2}{4} +\frac{e^2}{2} - \frac{1}{2} \\
&= \frac{3e^2}{4} - \frac{1}{4}
\end{align*}
Easy huh? Well, let us see why. That would be because $f$ is the inverse function to $x^x$ thus $f \left(x^x \right)=x$ and the rest follows naturally.
$$\mathcal{J}=\int_0^e f(x) \, {\rm d}x$$
(Euler Competition 2013)
Solution
Well for starters let me say that the special in the title does not stand for something WOW ! . It is just a category of exercises in computation of integrals. All we have to do is actually invoke the inverse function somehow.
Well,
\begin{align*}
\int_{0}^{e} f\left ( e^x \right ) \, {\rm d}x &\overset{e^x=y^y}{=\! =\! =\! =\!} \int_{1}^{e} y \left ( \ln y +1 \right ) \, {\rm d}y\\
&= \int_{1}^{e} y \ln y \, {\rm d}y + \int_{1}^{e} y \, {\rm d}y\\
&= \frac{1}{4} + \frac{e^2}{4} +\frac{e^2}{2} - \frac{1}{2} \\
&= \frac{3e^2}{4} - \frac{1}{4}
\end{align*}
Easy huh? Well, let us see why. That would be because $f$ is the inverse function to $x^x$ thus $f \left(x^x \right)=x$ and the rest follows naturally.
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