Question

If area bounded by region \({x, y)| |x^{2}-2| ≤ y ≤ x}\) is A, then 6A + 16\(\sqrt{2}\) is?

Answer 1

The given region can be described by the integral: \[ A = \int_1^{\sqrt{2}} \left( x - (2 - x^2) \right) dx + \int_{\sqrt{2}}^2 \left( x - (x^2 - 2) \right) dx \] Simplify the integrals: \[ A = \int_1^{\sqrt{2}} \left( x - 2 + x^2 \right) dx + \int_{\sqrt{2}}^2 \left( x - x^2 + 2 \right) dx \] Now, solve each integral: \[ \int_1^{\sqrt{2}} \left( x - 2 + x^2 \right) dx = \left[ \frac{x^2}{2} - 2x + \frac{x^3}{3} \right]_1^{\sqrt{2}} \] \[ = \left( \frac{2}{2} - 2\sqrt{2} + \frac{(\sqrt{2})^3}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \] \[ = \left( 1 - 2\sqrt{2} + \frac{2\sqrt{2}}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) \] \[ = 1 - 2\sqrt{2} + \frac{2\sqrt{2}}{3} + 2 - \frac{1}{2} - \frac{1}{3} \] Now the second integral: \[ \int_{\sqrt{2}}^2 \left( x - x^2 + 2 \right) dx = \left[ \frac{x^2}{2} - \frac{x^3}{3} + 2x \right]_{\sqrt{2}}^2 \] \[ = \left( \frac{4}{2} - \frac{8}{3} + 4 \right) - \left( \frac{2}{2} - \frac{2\sqrt{2}}{3} + 2\sqrt{2} \right) \] \[ = 2 - \frac{8}{3} + 4 - 1 + \frac{2\sqrt{2}}{3} - 2\sqrt{2} \] Now combine the results: \[ A = -4\sqrt{2} + \frac{4\sqrt{2}}{3} + \frac{7}{6} - \frac{8\sqrt{2}}{3} + \frac{9}{2} \] Now, calculate \( 6A + 16\sqrt{2} \): \[ 6A = -16\sqrt{2} + 27, \quad 6A + 16\sqrt{2} = 27. \] Thus, the correct answer is \( 27 \).