Let's break down the problem: **Given:** - The region is bounded by the line \( x = \frac{1}{2} \) and the curves \( y = x^2 \), \( y = \frac{1}{x} \). ### **Step 1: Find Points of Intersection** Set \( x^2 = \frac{1}{x} \): \[ x^3 = 1 \implies x = 1 \] At \( x = \frac{1}{2} \): - \( y_1 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \) - \( y_2 = \frac{1}{\frac{1}{2}} = 2 \) So, region is between \( x = \frac{1}{2} \) and \( x = 1 \), bounded by \( y = x^2 \) (below), \( y = \frac{1}{x} \) (above). ### **Step 2: Set up the Area Integral** For \( x \) from \( \frac{1}{2} \) to \( 1 \): \[ A_1 = \int_{x=\frac{1}{2}}^{x=1} \left( \frac{1}{x} - x^2 \right) dx \] But, we also need to check from \( x=1 \) to some upper bound if the region continues. For \( x > 1 \), \( y = \frac{1}{x} \) is **below** \( y = x^2 \), but above the \( x \)-axis, so this is not the bounded region described. But, let's check the options for the correct setup. ### **Step 3: Match with the Options** #### **(a) and (b)** Option (b) integrates from \( x=0 \) to \( x=2 \), but the region is only bounded between \( x=\frac{1}{2} \) and \( x=1 \) (from above analysis). #### **(c)** Option (c) \( \int_{0}^{1/2} \int_{0}^{y} (\sqrt{y} - \frac{1}{y}) dx\, dy \) is not matching our bounds. #### **(d) and (e)** Option (d) and (e) split the area into two regions. - For \( x \) from \( \frac{1}{4} \) to \( 1 \), \( y \) goes from \( x^2 \) up to \( \sqrt{y} \). - For \( x \) from \( 1 \) to \( 2 \), \( y \) goes from \( \frac{1}{x} \) up to \( \sqrt{y} \). But let's analyze option (e): #### **Option (e):** \[ \int_{\frac{1}{4}}^1 \int_{\frac{1}{2}}^{\sqrt{y}} dx\, dy + \int_{1}^2 \int_{\frac{1}{2}}^{\frac{1}{y}} dx\, dy \] Let's interpret: - The first integral: for \( y \) from \( 1/4 \) to \( 1 \), \( x \) goes from \( 1/2 \) to \( \sqrt{y} \). - The second: for \( y \) from \( 1 \) to \( 2 \), \( x \) goes from \( 1/2 \) to \( 1/y \). This matches the region bounded by \( x = 1/2 \), \( y = x^2 \), and \( y = 1/x \). ### **Final Answer** \[ \boxed{\text{(e)}} \] --- **Explanation:** The region is split into two parts: one where \( y = x^2 \) is below \( y = 1/x \) (for \( x \in [1/2, 1] \)), and the other for \( y \in [1, 2] \), where the bounds are given by \( x \in [1/2, 1/y] \). This setup is correctly described by option (e).

Let's analyze each option carefully and understand how they relate to the region bounded by the given curves and lines. --- ### **Given Curves and Lines:** - The vertical line: \( x = \frac{1}{2} \) - The curves: - \( y = x^2 \) (parabola opening upward) - \( y = \frac{1}{x} \) (hyperbola branch) --- ### **Step 1: Find intersection points of the curves** Set \( y = x^2 \) equal to \( y = \frac{1}{x} \): \[ x^2 = \frac{1}{x} \Rightarrow x^3 = 1 \Rightarrow x = 1 \] Corresponding \( y \): \[ y = (1)^2 = 1 \] At \( x = 1 \), both curves intersect at \( (1, 1) \). --- ### **Step 2: Find the bounds in \( x \)** - The region is bounded between \( x = \frac{1}{2} \) and \( x = 1 \), because: - For \( x \in [\frac{1}{2}, 1] \), the parabola \( y = x^2 \) lies **below** \( y = \frac{1}{x} \). - For \( x > 1 \), the line \( y = \frac{1}{x} \) becomes less than \( y = x^2 \), so the bounded region is only between \( x = 1/2 \) and \( x=1 \). --- ### **Step 3: Vertical slices to find area** - For each \( x \in [\frac{1}{2}, 1] \), the region is between: \[ y = x^2 \quad \text{(bottom)} \quad \text{and} \quad y = \frac{1}{x} \quad \text{(top)} \] - The **area** of the region can be expressed as: \[ A = \int_{x=\frac{1}{2}}^{x=1} \left( \frac{1}{x} - x^2 \right) dx \] --- ### **Step 4: Understanding the options** Now, let's analyze each option to see whether it correctly represents this region. --- ### **Option (a):** \[ \iint \text{ over some region } \quad \int_{z} \int_{y} dz\, dy \] - The integrand isn't specified here, but the notation suggests double integration over some limits. - Since the limits are not explicitly given, and the notation seems incomplete, **this is unlikely** the correct choice. --- ### **Option (b):** \[ \int_{x=0}^{2} \int_{z=1/2}^{\sqrt{y}} dy\, dx \] - It indicates integrating \( dy \) first from \( z=1/2 \) to \( \sqrt{y} \), but the limits are inconsistent: - The outer integral is in \( x \) from 0 to 2. - The inner integral seems to be in \( y \) from \( 1/2 \) to \( \sqrt{y} \), which doesn't make sense because the limits depend on \( y \). - Also, the order of integration and the bounds don't correspond to the region. **Conclusion:** Not matching the described region. --- ### **Option (c):** \[ \int_{0}^{1/2} \int_{0}^{y} \left( \sqrt{y} - \frac{1}{y} \right) dx\, dy \] - Limits: - Outer in \( y \) from 0 to \( 1/2 \) - Inner in \( x \) from 0 to \( y \) - The integrand is \( \sqrt{y} - \frac{1}{y} \), which doesn't relate directly to the curves \( y = x^2 \) and \( y=1/x \). - The bounds don't match the region between \( x=1/2 \) and \( x=1 \). **Conclusion:** Not appropriate. --- ### **Option (d):** \[ \int_{x=1/4}^1 \int_{z=1/2}^{\sqrt{y}} dx\, dy + \int_{x=1}^{2} \int_{z=1/2}^{1/y} dx\, dy \] - This appears to split the region into two parts: 1. For \( x \in [1/4, 1] \), \( y \) bounded by some functions involving \( z \) and \( y \). 2. For \( x \in [1, 2] \), similar bounds. - The bounds seem inconsistent with the region between \( y = x^2 \) and \( y=1/x \) for \( x \in [1/2, 1] \). **Conclusion:** Not matching the original region. --- ### **Option (e):** \[ \int_{y=1/4}^{1} \int_{x=1/2}^{\sqrt{y}} dx\, dy + \int_{y=1}^{2} \int_{x=1/2}^{1/y} dx\, dy \] - This is a **piecewise** double integral, split into two parts: 1. **Part 1:** \( y \in [1/4, 1] \) - \( x \) varies from \( 1/2 \) to \( \sqrt{y} \) 2. **Part 2:** \( y \in [1, 2] \) - \( x \) varies from \( 1/2 \) to \( 1/y \) - Let's interpret: - For \( y \in [1/4, 1] \), the upper bound in \( x \) is \( \sqrt{y} \). Since \( y \leq 1 \), \( \sqrt{y} \leq 1 \). - For \( y \in [1, 2] \), the upper bound in \( x \) is \( 1/y \), which decreases from 1 to 0.5. - These bounds approximate the region bounded by: - \( y = x^2 \) (which implies \( x = \sqrt{y} \)) - \( y = 1/x \) (which implies \( x = 1/y \)) - The line \( x = 1/2 \) - The bounds correctly describe the region between: - \( y = 1/4 \) and \( y=1 \), \( x \) from \( 1/2 \) up to \( \sqrt{y} \) - \( y=1 \) and \( y=2 \), \( x \) from \( 1/2 \) up to \( 1/y \) **Therefore**, **option (e)** accurately represents the region bounded between the curves \( y = x^2 \), \( y=1/x \), and the line \( x=1/2 \). --- ### **Final conclusion:** The correct integral setup matches **option (e)**. --- ### **Summary:** | Option | Description | Correct? | Explanation | |---------|--------------|----------|--------------| | (a) | Incomplete notation | No | Not explicitly described; unlikely | | (b) | Double integral with inconsistent bounds | No | Doesn't match region limits | | (c) | Double integral with limits from 0 to 1/2 | No | Doesn't match the region between the curves | | (d) | Split integral with bounds involving \( 1/4, 1, 2 \) | No | Not correctly representing the region | | **(e)** | Piecewise integral with bounds involving \( \sqrt{y} \) and \( 1/y \) | **Yes** | Correctly describes the bounded region between the curves | --- ## **Final answer:** \[ \boxed{\textbf{Option (e)}} \] This integral correctly captures the region bounded by \( x = \frac{1}{2} \), \( y = x^2 \), and \( y = \frac{1}{x} \).