To calculate the area of the shaded region, we will integrate the difference between the two curves over the interval where the shaded region is bounded.

The upper curve is $y = \sqrt{x + 2}$ and the lower curve is $y = \frac{1}{x + 1}$. The region is bounded between $x = 0$ and $x = 2$.

Step-by-Step Solution:

1. Set up the integral for the area: $\text{Area} = \int_0^2 \left[\sqrt{x + 2} - \frac{1}{x + 1}\right] dx$

2. Split the integral: $\text{Area} = \int_0^2 \sqrt{x + 2} \, dx - \int_0^2 \frac{1}{x + 1} \, dx$

3. Evaluate each integral:

   • For $\int \sqrt{x + 2} \, dx$: Use the substitution $u = x + 2$, so $du = dx$ and the bounds change to $u = 2$ to $u = 4$. $\int_0^2 \sqrt{x + 2} \, dx = \int_2^4 \sqrt{u} \, du = \int_2^4 u^{1/2} \, du$ Using the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1}$: $\int_2^4 u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_2^4 = \frac{2}{3}(4^{3/2}) - \frac{2}{3}(2^{3/2})$ Simplify: $4^{3/2} = \sqrt{4}^3 = 2^3 = 8, \quad 2^{3/2} = \sqrt{2}^3 = 2\sqrt{2}$ So: $\frac{2}{3}(8) - \frac{2}{3}(2\sqrt{2}) = \frac{16}{3} - \frac{4\sqrt{2}}{3}$

   • For $\int \frac{1}{x + 1} \, dx$: This is a standard logarithmic integral: $\int_0^2 \frac{1}{x + 1} \, dx = \ln|x + 1|_0^2 = \ln(3) - \ln(1)$ $= \ln(3)$

4. Combine the results: $\text{Area} = \left( \frac{16}{3} - \frac{4\sqrt{2}}{3} \right) - \ln(3)$

5. Simplify further if necessary: The final expression for the area is: $\text{Area} = \frac{16}{3} - \frac{4\sqrt{2}}{3} - \ln(3)$

Would you like a numerical approximation of this area?

Follow-Up Questions:

1. How do you determine the upper and lower curves for an integral of a shaded region?
2. Why is substitution helpful in evaluating $\int \sqrt{x+2} \, dx$?
3. What is the significance of logarithmic integration in this problem?
4. How can numerical methods be used to approximate the area if integration becomes complex?
5. How does the interval of integration affect the outcome?

Tip:

Always sketch the region if possible to visualize which curve is above or below and to