Question

The area (in sq. units) of the region \((x, y) : 0 \leq y \leq 2|x| + 1, 0 \leq y \leq x^2 + 1, |x| \leq 3\) is:

• A. \(\frac{64}{3}\)
• B. \(\frac{17}{3}\)
• C. \(\frac{32}{3}\)
• D. \(\frac{80}{3}\)

Hint:

Always check the limits and symmetry in region integrations to simplify the work.

Answer 1

The Correct Option is A

Step 1: Sketch the region. Identify intersections of lines and parabola within the range \( |x| \leq 3 \). 

Step 2: Solve for intersections. Solve \( 2|x| + 1 = x^2 + 1 \) for \( x \).2|x| = x^2 x = -2, 0, 2 (Only valid within the given range) 

Step 3: Integrate to find the area. Area =\( ∫_{-2}^{0} (x^2 + 1 - (2(-x) + 1)) dx + ∫_{0}^{2} (x^2 + 1 - (2x + 1)) dx = ∫_{-2}^{0} (x^2 - 2x) dx + ∫_{0}^{2} (x^2 - 2x) dx \)

Step 4: Calculate integrals. Area = \(2 × ∫_{0}^{2} (x^2 - 2x) dx = 2 × [ (x^3/3) - x^2 ]\)
 from 0 to 2 = 2 × [ (8/3) - 4 ] = 2 × [-4/3] = -8/3 
Total Area = 2 × | -8/3 | = 16/3