Question

The area bounded by the curves \( y = x^2 \) and \( y - 6 = -|x| \) is:

• A. \( \frac{37}{4} \)
• B. (no option shown)
• C. \( \frac{44}{3} \)
• D. \( \frac{38}{3} \)

Hint:

Use symmetry in area problems to simplify definite integrals involving even functions or absolute values.

Answer 1

The Correct Option is C

We are given the curves: \[ y = x^2 \quad \text{and} \quad y = 6 - |x| \]
To find the area between these curves, determine the points of intersection. Solve: \[ x^2 = 6 - |x| \]
Split into two cases: Case 1: \( x \geq 0 \Rightarrow |x| = x \): \[ x^2 = 6 - x \Rightarrow x^2 + x - 6 = 0 \Rightarrow (x + 3)(x - 2) = 0 \Rightarrow x = -3, 2 \]
Valid root: \( x = 2 \) Case 2: \( x \< 0 \Rightarrow |x| = -x \): \[ x^2 = 6 + x \Rightarrow x^2 - x - 6 = 0 \Rightarrow (x - 3)(x + 2) = 0 \Rightarrow x = 3, -2 \]
Valid root: \( x = -2 \) So, limits are from \( -2 \) to \( 2 \). Due to symmetry, compute area for \( [0, 2] \) and double it: \[ A = 2 \int_0^2 \left[(6 - x) - x^2 \right] dx = 2 \int_0^2 (6 - x - x^2) dx \]
\[ = 2 \left[6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_0^2 = 2 \left(12 - 2 - \frac{8}{3} \right) = 2 \left(10 - \frac{8}{3} \right) = 2 \cdot \frac{22}{3} = \frac{44}{3} \]