Question

The area of the region \( \{(x, y): 0 \leq y \leq x^2 + 1, \, 0 \leq y \leq x + 1, \, 0 \leq x \leq 2\ \) is:}

• A. \( \frac{23}{6} \)
• B. \( 2\sqrt{2} + 5 \)
• C. \( \frac{9}{2} \)
• D. None of these

Hint:

When calculating the area of a region enclosed by curves, split the region into smaller sections based on the intersection points of the curves for easier integration.

Answer 1

The Correct Option is A

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: \[ \boxed{\frac{23}{6}} \]