Find the area of the indicated region. (a) The region bounded by \( y=x^{2}-x-2 \) and \( y=2 x^{2}-4 x-6 \) (b) The region bounded by \( y=\sqrt{x}, y=1 \) and \( y=-x \)

Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{1} \left(\sqrt{x}-\left(-x\right)\right) dx\) - step1: Remove the parentheses: \(\int_{0}^{1} \left(\sqrt{x}+x\right) dx\) - step2: Evaluate the integral: \(\int \left(\sqrt{x}+x\right) dx\) - step3: Use properties of integrals: \(\int \sqrt{x} dx+\int x dx\) - step4: Evaluate the integral: \(\frac{2}{3}x^{\frac{3}{2}}+\int x dx\) - step5: Evaluate the integral: \(\frac{2}{3}x^{\frac{3}{2}}+\frac{x^{2}}{2}\) - step6: Return the limits: \(\left(\frac{2}{3}x^{\frac{3}{2}}+\frac{x^{2}}{2}\right)\bigg |_{0}^{1}\) - step7: Calculate the value: \(\frac{7}{6}\) Calculate the integral \( \int_{-1}^{3} ((2x^{2}-4x-6)-(x^{2}-x-2))dx \). Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-1}^{3} \left(\left(2x^{2}-4x-6\right)-\left(x^{2}-x-2\right)\right) dx\) - step1: Remove the parentheses: \(\int_{-1}^{3} \left(2x^{2}-4x-6-\left(x^{2}-x-2\right)\right) dx\) - step2: Subtract the terms: \(\int_{-1}^{3} \left(x^{2}-3x-4\right) dx\) - step3: Evaluate the integral: \(\int \left(x^{2}-3x-4\right) dx\) - step4: Use properties of integrals: \(\int x^{2} dx+\int -3x dx+\int -4 dx\) - step5: Evaluate the integral: \(\frac{x^{3}}{3}+\int -3x dx+\int -4 dx\) - step6: Evaluate the integral: \(\frac{x^{3}}{3}-\frac{3x^{2}}{2}+\int -4 dx\) - step7: Evaluate the integral: \(\frac{x^{3}}{3}-\frac{3x^{2}}{2}-4x\) - step8: Return the limits: \(\left(\frac{x^{3}}{3}-\frac{3x^{2}}{2}-4x\right)\bigg |_{-1}^{3}\) - step9: Calculate the value: \(-\frac{56}{3}\) Let's solve the problems step by step. ### (a) Area bounded by \( y = x^2 - x - 2 \) and \( y = 2x^2 - 4x - 6 \) 1. **Find the points of intersection**: We need to set the two equations equal to each other: \[ x^2 - x - 2 = 2x^2 - 4x - 6 \] Rearranging gives: \[ 0 = x^2 - 3x - 4 \] Factoring this equation: \[ 0 = (x - 4)(x + 1) \] Thus, the points of intersection are \( x = 4 \) and \( x = -1 \). 2. **Set up the integral for the area**: The area \( A \) between the curves from \( x = -1 \) to \( x = 4 \) is given by: \[ A = \int_{-1}^{4} \left( (2x^2 - 4x - 6) - (x^2 - x - 2) \right) dx \] 3. **Calculate the integral**: The result of the integral is: \[ A = -\frac{56}{3} \] Since area cannot be negative, we take the absolute value: \[ A = \frac{56}{3} \] ### (b) Area bounded by \( y = \sqrt{x} \), \( y = 1 \), and \( y = -x \) 1. **Find the points of intersection**: - For \( y = \sqrt{x} \) and \( y = 1 \): \[ \sqrt{x} = 1 \implies x = 1 \] - For \( y = \sqrt{x} \) and \( y = -x \): \[ \sqrt{x} = -x \implies x = 0 \quad (\text{since } \sqrt{x} \geq 0) \] - For \( y = 1 \) and \( y = -x \): \[ 1 = -x \implies x = -1 \] The relevant points of intersection for the area calculation are \( x = 0 \) and \( x = 1 \). 2. **Set up the integral for the area**: The area \( A \) between the curves from \( x = 0 \) to \( x = 1 \) is given by: \[ A = \int_{0}^{1} \left( \sqrt{x} - (-x) \right) dx \] 3. **Calculate the integral**: The result of the integral is: \[ A = \frac{7}{6} \] ### Final Answers: - (a) The area of the region bounded by \( y = x^2 - x - 2 \) and \( y = 2x^2 - 4x - 6 \) is \( \frac{56}{3} \). - (b) The area of the region bounded by \( y = \sqrt{x} \), \( y = 1 \), and \( y = -x \) is \( \frac{7}{6} \).