Tricky definite integral

Let $y=f(x)$ and $x=g(y)$ . Then $f$ and $g$ are the inverses of each other and $g(y)=ye^y$ (we consider the domain and range to be $[0,\infty)$ as the functions are bijective in this interval). Now observe the figure below which is the graph of $g(y)=ye^y$ so the reflection across the line $y=x$ is the graph of it's inverse.

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The marked rectangle has area $e$ so $\int_0^e f(x)~\text{d}x+\int_0^1 g(y)~\text{d}y=e\implies \int_0^e f(x)~\text{d}x=e-\int_0^1 g(y)~\text{d}y.$ Evaluating that $\int_0^1 g(y)~\text d y=\int_0^1 ye^y~\text d y=\left.e^y(y-1)\right|_0^1=1$ is a regular IBP exercise so left to the reader. Hence the value of our desired integral is $\boxed{e-1}.$

Note: $f\equiv W$ where $W$ is the Lambert $W$ -function.

Let $\displaystyle u=f(x) , du=f'(x) dx => dx=\frac{du}{f'(x)}$

Now differentiate implicitly our condition: $\displaystyle x=f(x)e^{f(x)} => 1= f'(x)e^{f(x)} +f(x)f'(x)e^{f(x)} , f'(x) = \frac{1}{(1+f(x))e^{f(x)}}$ .

Then : $\displaystyle dx= \frac{1}{f'(x)} du = (1+f(x))e^{f(x)} du = (1+u)e^{u} du$ .

Now we change the limits of integration : $f(0)=0 , f(e) = 1$ (Obviously) . Hence:

$\displaystyle \int_{0}^{e} f(x) dx = \int_0^1 u(1+u)e^u du = \int_0^1 (u^2+u)e^u du$

The last integral can be solved easily by integrating by parts (twice) to get the value $e-1$ from which we get our desired answer.

I am using parametric expression for $y$ and $x$ . Let $y(t) = \log(t)$ , then $x(t) =t \log(t)$ and $dx = \left(\log(t) +1\right) dt$ . Hence, we have $X= \int_{0}^{e} y dx =\int_{1}^{e} \log(t)\left(\log(t) +1\right) dt$ $X = t \left(\log^2 (t) +\log(t)+1\right)|^{e}_{1} = e -1$

Then, we have $[100X]=171$

substitute f(x) = z in 1st equation and you get the inverse function of f(x) as ze^z , now in the given integral substitute x= f^(-1) z and correspondingly change limits to 0 to 1 ( by inspection) and also put the derivative of inverse function multiplied by dz inplace of dx and use IBP to integrate