The given region is bounded by the following curves:
\[
y = x^2 + 1, \quad y = x + 1, \quad \text{and} \quad x = 0 \, \text{to} \, x = 2.
\]
Step 1: Break the region into subregions
For \( 0 \leq x \leq 1 \), the upper curve is \( y = x^2 + 1 \), while for \( 1 \leq x \leq 2 \), the upper curve is \( y = x + 1 \).
Step 2: Compute the total area
The total area \( A \) is the sum of two integrals:
\[
A = \int_{0}^{1} (x^2 + 1) \, dx + \int_{1}^{2} (x + 1) \, dx.
\]
1. First, evaluate \( \int_{0}^{1} (x^2 + 1) \, dx \):
\[
\int_{0}^{1} (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_{0}^{1} = \frac{1}{3}.
\]
2. Next, evaluate \( \int_{1}^{2} (x + 1) \, dx \):
\[
\int_{1}^{2} (x + 1) \, dx = \left[ \frac{x^2}{2} + x \right]_{1}^{2} = \left( \frac{4}{2} + 2 \right) - \left( \frac{1}{2} + 1 \right) = 3.5 = \frac{7}{2}.
\]
Step 3: Combine the results
Adding the two computed areas gives the total area:
\[
A = \frac{1}{3} + \frac{7}{2} = \frac{23}{6}.
\]
Final Answer:
\[
\boxed{\frac{23}{6}}
\]