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 \( x \).

\[ 2|x| = x^2 \]

\[ x = -2, 0, 2 \quad \text{(Only valid within the given range)} \]

Step 3: Integrate to find the area.

\[ \text{Area} = \int_{-2}^{0} (x^2 + 1 - (2(-x) + 1)) \, dx + \int_{0}^{2} (x^2 + 1 - (2x + 1)) \, dx \]

\[ = \int_{-2}^{0} (x^2 - 2x) \, dx + \int_{0}^{2} (x^2 - 2x) \, dx \]

Step 4: Calculate integrals.

\[ \text{Area} = 2 \times \int_{0}^{2} (x^2 - 2x) \, dx \]

\[ = 2 \times \left[ \frac{x^3}{3} - x^2 \right]_0^2 \]

\[ = 2 \times \left[ \frac{8}{3} - 4 \right] \]

\[ = 2 \times \left[ -\frac{4}{3} \right] = -\frac{8}{3} \]

\[ \text{Total Area} = 2 \times \left| -\frac{8}{3} \right| = \frac{16}{3} \]