Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a, b]$, is given by $\int_a^b [f(x) - g(x)] dx$.
• Intersection Points: To find where two curves intersect, set their equations equal to each other and solve for $x$ (and then $y$).
• Logarithm Properties: $\ln(a^b) = b\ln(a)$.

Step-by-Step Solution

Step 1: Understanding the Region and Identifying Bounding Curves

The problem asks for the area of the region defined by the inequalities:

1. $xy \le 8$
2. $1 \le y \le x^2$

We need to convert these inequalities into equations of boundary curves:

• $y = 1$: A horizontal line. Since $1 \le y$, the region is above this line.
• $y = x^2$: A parabola opening upwards with its vertex at the origin. Since $y \le x^2$, the region is below the parabola.
• $y = 8/x$: A rectangular hyperbola. Since $xy \le 8$, the region is below the hyperbola (for $x > 0$).

Why this step is taken: Visualizing the region is essential. Sketching the curves helps identify the upper and lower bounds for integration.

Let's sketch these curves:

• Draw the line $y=1$.
• Draw the parabola $y=x^2$.
• Draw the hyperbola $y=8/x$. Since $y \ge 1$, we are only interested in the first quadrant ($x>0, y>0$).

The region we are looking for is above $y=1$, below $y=x^2$, and below $y=8/x$.

Step 2: Finding Intersection Points

To determine the limits of integration, we need to find where these curves intersect each other.

• Intersection of $y=1$ and $y=x^2$: Set $x^2 = 1 \implies x = \pm 1$. Since we're in the first quadrant, we take $x=1$. Thus $y=1$. Point A: $(1,1)$.

• Intersection of $y=1$ and $y=8/x$: Set $8/x = 1 \implies x = 8$. Thus $y=1$. Point B: $(8,1)$.

• Intersection of $y=x^2$ and $y=8/x$: Set $x^2 = 8/x \implies x^3 = 8 \implies x = 2$. Substituting $x=2$ into $y=x^2$ gives $y = 2^2 = 4$. Point C: $(2,4)$.

Why this step is taken: Intersection points define the boundaries of the sub-regions where the upper and lower curves remain consistent. The $x$-coordinates will be our limits of integration.

Step 3: Setting Up the Integral(s)

From our sketch and the intersection points, we observe that the upper bounding curve changes at $x=2$.

• For $x$ from $1$ to $2$: The region is bounded above by $y=x^2$ and below by $y=1$.
• For $x$ from $2$ to $8$: The region is bounded above by $y=8/x$ and below by $y=1$.

Therefore, we need to split the total area into two integrals: Total Area $A = A_1 + A_2$

• Area $A_1$ (from $x=1$ to $x=2$): The upper curve is $f(x) = x^2$. The lower curve is $g(x) = 1$. $A_1 = \int_1^2 (x^2 - 1) dx$

• Area $A_2$ (from $x=2$ to $x=8$): The upper curve is $f(x) = 8/x$. The lower curve is $g(x) = 1$. $A_2 = \int_2^8 \left(\frac{8}{x} - 1\right) dx$

Why this step is taken: This is where we translate the visual understanding into the mathematical setup for integration. It's crucial to correctly identify which function is "upper" and which is "lower" in each interval.

Step 4: Evaluating the Integrals

Now, we evaluate each definite integral:

• For $A_1$: $A_1 = \int_1^2 (x^2 - 1) dx = \left[ \frac{x^3}{3} - x \right]_1^2$ $A_1 = \left( \frac{2^3}{3} - 2 \right) - \left( \frac{1^3}{3} - 1 \right)$ $A_1 = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 1 \right)$ $A_1 = \left( \frac{8-6}{3} \right) - \left( \frac{1-3}{3} \right)$ $A_1 = \frac{2}{3} - \left( -\frac{2}{3} \right) = \frac{2}{3} + \frac{2}{3} = \frac{4}{3}$

• For $A_2$: $A_2 = \int_2^8 \left(\frac{8}{x} - 1\right) dx = \left[ 8 \ln|x| - x \right]_2^8$ $A_2 = (8 \ln|8| - 8) - (8 \ln|2| - 2)$ $A_2 = 8 \ln 8 - 8 - 8 \ln 2 + 2$ $A_2 = 8 \ln 8 - 8 \ln 2 - 6$

Why this step is taken: This is the core calculation. Accurate integration and careful substitution of limits are essential.

Step 5: Simplifying the Result

Now, we add $A_1$ and $A_2$ and simplify using logarithm properties. Recall that $\ln 8 = \ln (2^3) = 3 \ln 2$.

$A_2 = 8(3 \ln 2) - 8 \ln 2 - 6$ $A_2 = 24 \ln 2 - 8 \ln 2 - 6$ $A_2 = 16 \ln 2 - 6$

Now, the total area $A = A_1 + A_2$: $A = \frac{4}{3} + (16 \ln 2 - 6)$ $A = 16 \ln 2 + \frac{4}{3} - \frac{18}{3}$ $A = 16 \ln 2 - \frac{14}{3}$

Why this step is taken: Final simplification ensures the answer is in its most concise form and allows for easy comparison with the given options.

Common Mistakes & Tips

• Forgetting to split the integral when the upper or lower boundary function changes.
• Incorrectly calculating intersection points, leading to wrong limits of integration.
• Not sketching the region, making it difficult to identify the correct upper and lower functions.

Summary

This problem required finding the area of a region bounded by a line, a parabola, and a hyperbola. The key was to correctly identify the region, find the intersection points of the curves, set up the appropriate integrals by splitting the region into sub-regions based on the bounding curves, evaluate the integrals, and simplify the result. The total area is $16 \ln 2 - \frac{14}{3}$.

Final Answer

The final answer is $\boxed{16{\log _e}2 - {{14} \over 3}}$, which corresponds to option (A).