Key Concepts and Formulas

• Area between curves: If $f(x) \ge g(x)$ on $[a, b]$, the area is $\int_a^b [f(x) - g(x)] \, dx$.
• Integration of $\frac{1}{x}$: $\int \frac{1}{x} \, dx = \ln|x| + C$.
• Logarithm properties: $\ln a - \ln b = \ln(\frac{a}{b})$, $\ln(a^b) = b \ln a$.

Step-by-Step Solution

Step 1: Express the equations in the form $y = f(x)$

We need to rewrite both equations to isolate $y$ as a function of $x$. This allows us to easily determine which curve is "above" the other and set up the integral correctly.

• Equation 1: $xy + 4y = 16$ Factor out $y$: $y(x + 4) = 16$ Divide by $(x + 4)$: $y = \frac{16}{x + 4}$. Let $y_1(x) = \frac{16}{x+4}$.

• Equation 2: $x + y = 6$ Subtract $x$ from both sides: $y = 6 - x$. Let $y_2(x) = 6 - x$.

Step 2: Find the points of intersection

The intersection points occur where the $y$-values of the two curves are equal. Finding these points will give us the limits of integration.

Set $y_1(x) = y_2(x)$: $\frac{16}{x + 4} = 6 - x$ Multiply both sides by $(x + 4)$: $16 = (6 - x)(x + 4)$ Expand the right side: $16 = 24 + 6x - 4x - x^2$ $16 = 24 + 2x - x^2$ Rearrange to form a quadratic equation: $x^2 - 2x - 8 = 0$ Factor the quadratic: $(x - 4)(x + 2) = 0$ Solve for $x$: $x = 4 \text{ or } x = -2$ So, the limits of integration are $x = -2$ and $x = 4$.

Step 3: Determine which curve is on top

We need to determine which function, $y_1(x)$ or $y_2(x)$, has larger values in the interval $[-2, 4]$. Choose a test point within the interval, say $x = 0$. $y_1(0) = \frac{16}{0 + 4} = 4$ $y_2(0) = 6 - 0 = 6$ Since $y_2(0) > y_1(0)$, the line $y_2(x) = 6 - x$ is above the hyperbola $y_1(x) = \frac{16}{x + 4}$ on the interval $[-2, 4]$.

Step 4: Set up the definite integral

The area between the curves is given by the integral of the difference between the upper and lower functions: $A = \int_{-2}^{4} [y_2(x) - y_1(x)] \, dx = \int_{-2}^{4} \left[(6 - x) - \frac{16}{x + 4}\right] \, dx$

Step 5: Evaluate the definite integral

$A = \int_{-2}^{4} \left(6 - x - \frac{16}{x + 4}\right) \, dx$ Integrate each term separately: $A = \left[6x - \frac{x^2}{2} - 16\ln|x + 4|\right]_{-2}^{4}$ Evaluate at the upper and lower limits: $A = \left(6(4) - \frac{4^2}{2} - 16\ln|4 + 4|\right) - \left(6(-2) - \frac{(-2)^2}{2} - 16\ln|-2 + 4|\right)$ $A = \left(24 - 8 - 16\ln 8\right) - \left(-12 - 2 - 16\ln 2\right)$ $A = \left(16 - 16\ln 8\right) - \left(-14 - 16\ln 2\right)$ $A = 16 - 16\ln 8 + 14 + 16\ln 2$ $A = 30 - 16\ln 8 + 16\ln 2$ $A = 30 - 16\ln(2^3) + 16\ln 2$ $A = 30 - 16(3\ln 2) + 16\ln 2$ $A = 30 - 48\ln 2 + 16\ln 2$ $A = 30 - 32\ln 2$

Common Mistakes & Tips

• Remember to check which function is greater on the interval of integration. Switching the order will result in the negative of the correct area.
• Be careful with the signs when evaluating the definite integral at the limits of integration.
• Utilize logarithm properties to simplify the final answer and match it to the given options.

Summary

We found the area enclosed by the curves $xy + 4y = 16$ and $x + y = 6$ by expressing both equations in the form $y = f(x)$, finding their intersection points to determine the limits of integration, determining which curve was on top, and then evaluating the definite integral. The enclosed area is $30 - 32\ln 2$.

Final Answer

The final answer is \boxed{30-32 \log _{\mathrm{e}} 2}, which corresponds to option (C).