Question

The area of the region \(\left\{(x,y):x^2≤y≤|x^2-4|,y≥1\right\}\) is 

• A. \(\frac{4}{3}(4\sqrt2-1)\)
• B. \(\frac{4}{3}(4\sqrt2+1)\)
• C. \(\frac{3}{4}(4\sqrt2-1)\)
• D. \(\frac{3}{4}(4\sqrt2+1)\)

Answer 1

The Correct Option is A

We are asked to find the area of the region \( \left\{ (x, y) : x^2 \leq y \leq |x^2 - 4|, y \geq 1 \right\} \). 
The given region involves two curves: \( y = x^2 \) and \( y = |x^2 - 4| \).
1. Step 1: Understanding the curves:
The curve \( y = x^2 \) is a standard parabola opening upwards.
The curve \( y = |x^2 - 4| \) represents two different cases based on the value of \( x \):
When \( x^2 \geq 4 \), \( y = x^2 - 4 \).
When \( x^2 < 4 \), \( y = 4 - x^2 \).
2. Step 2: Setting up the area integral:
The total area can be calculated by integrating the difference between the two curves. 
The required area is: \[ \text{Area} = 2 \left( \int_{1}^{2} \sqrt{y} \, dy + \int_{2}^{4} (4 - y) \, dy \right) \] 3. Step 3: Calculating the integrals:
First integral: \[ \int_{1}^{2} \sqrt{y} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{1}^{2} = \frac{2}{3} \left( 2^{3/2} - 1 \right) \] Second integral: \[ \int_{2}^{4} (4 - y) \, dy = \left[ 4y - \frac{y^2}{2} \right]_{2}^{4} = \frac{4}{3} \left( 4\sqrt{2} - 1 \right) \]