Chapter 5: Integrals - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises 5.6 - Page 304: 73

$1$

Step 1. See figure. We get the intersections as $(1,1)$ and $(2,\frac{1}{4})$ for the enclosed region of concern. Step 2. The enclosed area among the three functions can be found as $A=\int_0^1 (x)dx+\int_1^2 (\frac{1}{x^2})dx =\frac{1}{2}x^2|_{0}^1- \frac{1}{x}|_{1}^2 =\frac{1}{2}(1)^2- \frac{1}{2}+ \frac{1}{1}=1$