## Calculus: Early Transcendentals 8th Edition

$$A = \ln (2) - \frac{1}{2}$$
The area enclosed between two curves $y = f(x)$ and $y = g(x)$ over $[a,b]$ such that $f(x) \geq g(x)$ is $$\displaystyle\int_{a}^{b} f(x) - g(x) dx$$ From the graph we can see that $\displaystyle\frac{1}{x} \geq \frac{1}{x^{2}}$ on the interval $[1,2]$ Therefore, the area between the two curves is $$A = \displaystyle\int_{1}^{2} \frac{1}{x} - \frac{1}{x^{2}} dx$$ $$A = \left[ \ln(x)+ \frac{1}{x} \right]^{2}_1$$ $$A = \bigg[ \ln (s) + \frac{1}{2} \bigg] - \bigg[ \ln(1) + \frac{1}{1} \bigg]$$ $$A = \ln (2) +\frac{1}{2}-1$$ $$A = \ln (2) - \frac{1}{2}$$