18. Find the area of the closed region bounded by the two curves \( y=x^{3}+2 x^{2}-3 x \) and \( y=2 x^{2}+x \) A 2 \( \begin{array}{lll}\text { B } 3 & \text { D } 6 & \text { E } 8\end{array} \)

To find the area of the closed region bounded by the two curves \( y=x^{3}+2x^{2}-3x \) and \( y=2x^{2}+x \), we need to follow these steps: 1. Find the points of intersection of the two curves. 2. Determine the limits of integration based on the points of intersection. 3. Set up the integral to find the area between the curves. 4. Calculate the integral to find the area. Let's start by finding the points of intersection of the two curves. We need to solve the equation \( x^{3}+2x^{2}-3x = 2x^{2}+x \) to find the points of intersection. Solve the equation by following steps: - step0: Solve for \(x\): \(x^{3}+2x^{2}-3x=2x^{2}+x\) - step1: Cancel equal terms: \(x^{3}-3x=x\) - step2: Move the expression to the left side: \(x^{3}-3x-x=0\) - step3: Subtract the terms: \(x^{3}-4x=0\) - step4: Factor the expression: \(x\left(x^{2}-4\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x=0\\&x^{2}-4=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=0\\&x=2\\&x=-2\end{align}\) - step7: Rewrite: \(x_{1}=-2,x_{2}=0,x_{3}=2\) The points of intersection of the two curves are \( x=-2, x=0, \) and \( x=2 \). Next, we need to determine the limits of integration based on these points of intersection. The limits of integration will be from \( x=-2 \) to \( x=2 \) since these are the points where the curves intersect. Now, we will set up the integral to find the area between the curves. The area between the curves can be found by integrating the absolute difference between the two functions over the interval \([-2, 2]\). The integral to find the area is given by: \[ \text{Area} = \int_{-2}^{2} |(x^{3}+2x^{2}-3x) - (2x^{2}+x)| dx \] Let's calculate this integral to find the area. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{-2}^{2} \left|\left(x^{3}+2x^{2}-3x\right)-\left(2x^{2}+x\right)\right| dx\) - step1: Remove the parentheses: \(\int_{-2}^{2} \left|x^{3}+2x^{2}-3x-\left(2x^{2}+x\right)\right| dx\) - step2: Subtract the terms: \(\int_{-2}^{2} \left|x^{3}-4x\right| dx\) - step3: Rewrite the expression: \(\int_{-2}^{0} x^{3}-4x dx+\int_{0}^{2} -x^{3}+4x dx\) - step4: Evaluate the integral: \(4+\int_{0}^{2} -x^{3}+4x dx\) - step5: Evaluate the integral: \(4+4\) - step6: Calculate: \(8\) The area of the closed region bounded by the two curves \( y=x^{3}+2x^{2}-3x \) and \( y=2x^{2}+x \) is 8 square units. Therefore, the correct answer is option E: 8.