Area — Question 2

Question 2 - Solution

On the interval $0<x\le 1$, the curves intersect where $$\ln x=x-1,$$ which occurs at $$x=1.$$

For $0<x<1$, we have $$x-1 \ge \ln x,$$ so the area between the curves is $$A=\int_{0}^{1}\bigl((x-1)-\ln x\bigr)\,dx.$$

Compute each term: $$\int_{0}^{1}(x-1)\,dx=\left[\frac{x^2}{2}-x\right]_0^1=-\frac12,$$ $$\int_{0}^{1}\ln x\,dx=\left[x\ln x-x\right]_0^1=-1.$$

Therefore, $$A=-\frac12-(-1)=\frac12.$$

$$\boxed{\frac12}$$