for each problem, find the area of the region enclosed by the curves. you may use the provided graph to…

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n6) $y = 2\\sqrt3{x^2}$\n$y = x$

for each problem, find the area of the region enclosed by the curves. you may use the provided graph to sketch the curves and shade the enclosed region.\n\n6) $y = 2\\sqrt3{x^2}$\n$y = x$

Answer

Explanation:

Step1: Find the intersection points of the curves

Set the equations equal to each other: $2\sqrt[3]{x^{2}} = x$. Cube both sides: $8x^{2} = x^{3}$.

Step2: Solve for the x-coordinates

Rearrange the equation: $x^{3} - 8x^{2} = 0 \Rightarrow x^{2}(x - 8) = 0$. The solutions are $x = 0$ and $x = 8$.

Step3: Set up the definite integral

Identify the upper curve. For $0 < x < 8$, $2x^{2/3} > x$. The area $A$ is: $$A = \int_{0}^{8} (2x^{2/3} - x) , dx$$

Step4: Evaluate the integral

Find the antiderivative and apply the limits: $$A = \left[ 2 \cdot \frac{x^{5/3}}{5/3} - \frac{x^{2}}{2} \right]{0}^{8} = \left[ \frac{6}{5}x^{5/3} - \frac{x^{2}}{2} \right]{0}^{8}$$

Step5: Calculate the final value

Substitute $x = 8$: $$A = \frac{6}{5}(8)^{5/3} - \frac{8^{2}}{2} = \frac{6}{5}(32) - \frac{64}{2} = \frac{192}{5} - 32 = 38.4 - 32 = 6.4$$

Answer:

6.4