What is the area of the region enclosed by Ox, the curve $ y=\ln(x) $ and its tangent?

Question

First of all I determined the tangent. Let $d1$ be $y=\ln(x)$ and its tangent $dtan$. $dtan$ passes through $ (0,0) $ so it should be tangent to $d1$ in $ (a,b) $. I used the formula $y-\ln(a)=\frac{1}{a}\left(x-a\right)$ and replaced $x$ and $y$ with $0$. Like this, I determined that $a=e$ and $b=1$. Now, $dtan: y=\frac{x}{e}$. $\space\ln(x)$ and $\frac{x}{e}$ are both continuous so they can be integrated. Since $\ln(x)\lt\frac{x}{e}$, the area should be $\int _0^e\:\frac{x}{e}-\ln\left(x\right)dx$ which is $\frac{e}{2}$. The problem is that this is not the answer. It should be $\frac{e}{2} -1$. Have I missed something important?

Answer 1

All calculations are correct except the integral limits. If you plot the curves, you'll see that you have to split the integral in two parts as the region is limited by three curves. The integral has to be from $0$ to $1$ only for $x/e$ and from $1$ to $e$ for the one you wrote with $x/e-\ln x$. The way you've stated the limits, you are integrating the logarithm for all its negative values.