To find the area of the shaded region between the curves $y = \sqrt{x + 2}$ and $y = \frac{1}{x + 1}$ from $x = 0$ to $x = 2$, we'll follow these steps:

Step 1: Set up the integral

The area $A$ of the region between two curves $y_1(x)$ and $y_2(x)$ from $x = a$ to $x = b$ is given by: $A = \int_{a}^{b} \left[ y_1(x) - y_2(x) \right] \, dx$ Here, $y_1(x) = \sqrt{x + 2}$ (the upper curve) and $y_2(x) = \frac{1}{x + 1}$ (the lower curve). The bounds are from $x = 0$ to $x = 2$.

Step 2: Evaluate the integral

The integral becomes: $A = \int_{0}^{2} \left( \sqrt{x + 2} - \frac{1}{x + 1} \right) \, dx$

Let's compute this integral step-by-step.

1. Integral of $\sqrt{x + 2}$:

   • Use the substitution $u = x + 2$, so $du = dx$. $\int \sqrt{x + 2} \, dx = \int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C = \frac{2}{3} (x + 2)^{3/2}$

2. Integral of $\frac{1}{x + 1}$:

   • This is a standard integral: $\int \frac{1}{x + 1} \, dx = \ln|x + 1| + C$

So, the area is: $A = \left[ \frac{2}{3} (x + 2)^{3/2} - \ln|x + 1| \right]_{0}^{2}$

Step 3: Compute the definite integral

Substitute the bounds $x = 2$ and $x = 0$ into the expression:

1. At $x = 2$: $\frac{2}{3} (2 + 2)^{3/2} - \ln(2 + 1) = \frac{2}{3} \cdot 4^{3/2} - \ln(3) = \frac{2}{3} \cdot 8 - \ln(3) = \frac{16}{3} - \ln(3)$

2. At $x = 0$: $\frac{2}{3} (0 + 2)^{3/2} - \ln(0 + 1) = \frac{2}{3} \cdot 2^{3/2} - \ln(1) = \frac{2 \cdot 2\sqrt{2}}{3} - 0 = \frac{4\sqrt{2}}{3}$

So, the area is: $A = \left( \frac{16}{3} - \ln(3) \right) - \frac{4\sqrt{2}}{3}$ $A = \frac{16}{3} - \frac{4\sqrt{2}}{3} - \ln(3)$

Step 4: Simplify the final answer

Combine the terms to get the final area: $A = \frac{16 - 4\sqrt{2}}{3} - \ln(3)$ This expression gives the exact area of the shaded region.

Would you like further details on any part of this process, or help with something else?

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Related Questions:

1. How would the area change if the region was between $x = 1$ and $x = 2$ instead of $x = 0$ to $x = 2$?
2. Can you determine where the two curves intersect?
3. What is the maximum distance between the two curves in the interval $x = 0$ to $x = 2$?
4. How do you determine which curve is the upper one without plotting?
5. What happens to the area if the upper and lower curves are switched?

Tip:

Always check the intersection points between curves to accurately set the limits of integration when finding areas between curves.