Let's solve the problem step by step: Given functions: - \( y = 2x \) - \( y = 4 \) - \( y = \frac{1}{4}x^2 \) We are to find the area of the region enclosed by these curves. --- ### 1. **Find Points of Intersection** #### **(a) Intersection of \(y = 2x\) and \(y = 4\):** Set \( 2x = 4 \implies x = 2 \). #### **(b) Intersection of \(y = \frac{1}{4}x^2\) and \(y = 4\):** Set \( \frac{1}{4}x^2 = 4 \implies x^2 = 16 \implies x = \pm 4 \). #### **(c) Intersection of \(y = 2x\) and \(y = \frac{1}{4}x^2\):** Set \( 2x = \frac{1}{4}x^2 \implies x^2 - 8x = 0 \implies x(x-8)=0 \implies x=0 \) or \( x=8 \). --- ### 2. **Determine Bounds for Integration** The region is enclosed between: - Left: \(x = 0\) (intersection of \(y = 2x\) and \(y = \frac{1}{4}x^2\)) - Right: \(x = 2\) (intersection of \(y = 2x\) and \(y = 4\)) - Top: \(y = 4\) - Bottom: \(y = \frac{1}{4}x^2\) But let's confirm the relevant bounds: - From \(x=0\) to \(x=2\), the upper bound is \(y=2x\), and the lower bound is \(y=\frac{1}{4}x^2\). - From \(x=2\) to \(x=4\), the upper bound is \(y=4\), and the lower bound is \(y=\frac{1}{4}x^2\). --- ### 3. **Set Up the Area Integral** - **From \(x = 0\) to \(x = 2\):** \[ A_1 = \int_{0}^{2} [2x - \frac{1}{4}x^2]\, dx \] - **From \(x = 2\) to \(x = 4\):** \[ A_2 = \int_{2}^{4} [4 - \frac{1}{4}x^2]\, dx \] --- ### 4. **Evaluate the Integrals** #### **(a) \(A_1:\)** \[ A_1 = \int_{0}^{2} (2x - \frac{1}{4}x^2) dx \] \[ = \left[ x^2 - \frac{1}{12}x^3 \right]_{0}^{2} \] Plug in the values: \[ = \left( (2)^2 - \frac{1}{12}(2)^3 \right) - (0 - 0) \] \[ = (4 - \frac{8}{12}) \] \[ = 4 - \frac{2}{3} \] \[ = \frac{12}{3} - \frac{2}{3} = \frac{10}{3} \] --- #### **(b) \(A_2:\)** \[ A_2 = \int_{2}^{4} (4 - \frac{1}{4}x^2) dx \] \[ = \left[ 4x - \frac{1}{12}x^3 \right]_{2}^{4} \] Plug in the values: \[ = \left( 4(4) - \frac{1}{12}(4)^3 \right) - \left( 4(2) - \frac{1}{12}(2)^3 \right) \] \[ = (16 - \frac{64}{12}) - (8 - \frac{8}{12}) \] \[ = (16 - \frac{16}{3}) - (8 - \frac{2}{3}) \] \[ = \left( \frac{48}{3} - \frac{16}{3} \right) - \left( \frac{24}{3} - \frac{2}{3} \right) \] \[ = \frac{32}{3} - \frac{22}{3} \] \[ = \frac{10}{3} \] --- ### 5. **Total Area** \[ A = A_1 + A_2 = \frac{10}{3} + \frac{10}{3} = \frac{20}{3} \] --- ## **Final Answer** \[ \boxed{\frac{20}{3}} \] This is the area of the shaded region enclosed by the given functions.

The enclosed region is found by integrating the difference between the upper and lower curves over the relevant intervals, which correspond to their points of intersection.

The total area is obtained by summing the integrals of the difference between the top and bottom functions over the intervals where they form the enclosed region.