Let's break down the **area of the shaded region** step by step: ### 1. **Find the intersection points** #### **Between \( y = 7x \) and \( y = 44 - x^2 \):** Set \( 7x = 44 - x^2 \): \[ x^2 + 7x - 44 = 0 \] \[ (x + 11)(x - 4) = 0 \] So, \( x = -11 \) or \( x = 4 \). Since we are in the first quadrant, **use \( x = 4 \)**. --- #### **Between \( y = \frac{4}{3}x \) and \( y = 44 - x^2 \):** Set \( \frac{4}{3}x = 44 - x^2 \): \[ x^2 + \frac{4}{3}x - 44 = 0 \] Multiply both sides by 3: \[ 3x^2 + 4x - 132 = 0 \] \[ (x + 6)(3x - 22) = 0 \] So, \( x = -6 \) or \( x = \frac{22}{3} \). Since we are in the first quadrant, **use \( x = \frac{22}{3} \)**. --- #### **Both lines and the parabola intersect at \( x = 0 \):** At \( x=0 \): \( y = 0 \) for both lines and \( y = 44 \) for the parabola. But for area, we only consider from \( x = 0 \). --- ### 2. **Set up the integral** The shaded region is bounded above by \( y = 44 - x^2 \), below by \( y = 7x \) on the left and \( y = \frac{4}{3}x \) on the right. #### The limits: - From \( x = 0 \) to \( x = 4 \): lower bound is \( y = 7x \) - From \( x = 4 \) to \( x = \frac{22}{3} \): lower bound is \( y = \frac{4}{3}x \) #### **Area is:** \[ \text{Area} = \int_0^4 \left[ (44 - x^2) - 7x \right] dx + \int_4^{\frac{22}{3}} \left[ (44 - x^2) - \frac{4}{3}x \right] dx \] --- ### 3. **Calculate the integrals** #### **First integral:** \[ \int_0^4 (44 - x^2 - 7x) dx = \int_0^4 (44 - 7x - x^2) dx \] \[ = \left[ 44x - \frac{7}{2}x^2 - \frac{1}{3}x^3 \right]_0^4 \] Plug in \( x = 4 \): - \( 44 \times 4 = 176 \) - \( \frac{7}{2} \times 16 = 56 \) - \( \frac{1}{3} \times 64 = \frac{64}{3} \) So, \[ = 176 - 56 - \frac{64}{3} = 120 - \frac{64}{3} = \frac{360 - 64}{3} = \frac{296}{3} \] --- #### **Second integral:** \[ \int_4^{\frac{22}{3}} (44 - x^2 - \frac{4}{3}x) dx = \left[ 44x - \frac{1}{3}x^3 - \frac{2}{3}x^2 \right]_4^{\frac{22}{3}} \] Calculate at \( x = \frac{22}{3} \): - \( 44 \times \frac{22}{3} = \frac{968}{3} \) - \( \left(\frac{22}{3}\right)^3 = \frac{10648}{27} \implies \frac{1}{3} \times \frac{10648}{27} = \frac{10648}{81} \) - \( \left(\frac{22}{3}\right)^2 = \frac{484}{9} \implies \frac{2}{3} \times \frac{484}{9} = \frac{968}{27} \) So, \[ \text{At } x = \frac{22}{3}: \frac{968}{3} - \frac{10648}{81} - \frac{968}{27} \] At \( x = 4 \): - \( 44 \times 4 = 176 \) - \( \frac{1}{3} \times 64 = \frac{64}{3} \) - \( \frac{2}{3} \times 16 = \frac{32}{3} \) So, \[ 176 - \frac{64}{3} - \frac{32}{3} = 176 - \frac{96}{3} = 176 - 32 = 144 \] The definite integral: \[ \left[ \frac{968}{3} - \frac{10648}{81} - \frac{968}{27} \right] - 144 \] Let's get a common denominator for the first part: - \(\frac{968}{3} = \frac{26136}{81}\) - \(\frac{968}{27} = \frac{2904}{81}\) So, \[ \frac{26136}{81} - \frac{10648}{81} - \frac{2904}{81} = \frac{26136 - 10648 - 2904}{81} = \frac{12584}{81} \] Minus 144: \[ \frac{12584}{81} - 144 = \frac{12584 - (144 \times 81)}{81} = \frac{12584 - 11664}{81} = \frac{920}{81} \] --- ### 4. **Total Area** \[ \text{Area} = \frac{296}{3} + \frac{920}{81} \] Make common denominator: - \(\frac{296}{3} = \frac{7992}{81}\) So, \[ \text{Area} = \frac{7992 + 920}{81} = \frac{8912}{81} \] --- ## **Final Answer** \[ \boxed{\frac{8912}{81}} \] This is the exact area of the shaded region.

Let's carefully recompute the area of the shaded region step by step, ensuring accuracy. --- ### Step 1: Find intersection points **Between \( y = 7x \) and \( y = 44 - x^2 \):** Set \(7x = 44 - x^2\): \[ x^2 + 7x - 44 = 0 \] Factor: \[ (x + 11)(x - 4) = 0 \] Solutions: \[ x = -11 \quad \text{or} \quad x=4 \] Since we're interested in the first quadrant, the relevant intersection is at: \[ x=4 \] Corresponding \( y \): \[ y=7 \times 4 = 28 \] --- **Between \( y = \frac{4}{3}x \) and \( y = 44 - x^2 \):** Set \( \frac{4}{3}x = 44 - x^2 \): \[ x^2 + \frac{4}{3}x -44=0 \] Multiply through by 3: \[ 3x^2 + 4x -132=0 \] Factor or use quadratic formula: \[ x = \frac{-4 \pm \sqrt{(4)^2 - 4 \times 3 \times (-132)}}{2 \times 3} \] Calculate discriminant: \[ 16 + 4 \times 3 \times 132 = 16 + 4 \times 396 = 16 + 1584 = 1600 \] Square root: \[ \sqrt{1600} = 40 \] Solutions: \[ x = \frac{-4 \pm 40}{6} \] - \( x= \frac{-4 + 40}{6} = \frac{36}{6} =6 \) - \( x= \frac{-4 - 40}{6} = \frac{-44}{6} = -\frac{22}{3} \) In the first quadrant, only positive \( x \): \[ x=6 \] Corresponding \( y \): \[ y= \frac{4}{3} \times 6=8 \] --- ### Step 2: Determine the bounds of integration The relevant intersection points are at: - \( x=0 \), \( y=0 \), where all three curves meet - \( x=4 \), \( y=28 \) - \( x=6 \), \( y=8 \) The shaded region is bounded between: - \( x=0 \) to \( x=4 \): between \( y=7x \) and \( y=44 - x^2 \) - \( x=4 \) to \( x=6 \): between \( y=\frac{4}{3}x \) and \( y=44 - x^2 \) --- ### Step 3: Set up the integrals **Area \(A\):** \[ A = \int_{0}^{4} \left[ (44 - x^2) - 7x \right] dx + \int_{4}^{6} \left[ (44 - x^2) - \frac{4}{3}x \right] dx \] --- ### Step 4: Compute the first integral \[ I_1 = \int_{0}^{4} (44 - x^2 - 7x) dx \] Antiderivative: \[ 44x - \frac{x^3}{3} - \frac{7}{2}x^2 \] Evaluate from 0 to 4: At \( x=4 \): \[ 44 \times 4 - \frac{64}{3} - \frac{7}{2} \times 16 = 176 - \frac{64}{3} - 56 \] Simplify: \[ (176 - 56) - \frac{64}{3} = 120 - \frac{64}{3} = \frac{360}{3} - \frac{64}{3} = \frac{296}{3} \] At \( x=0 \), integral is 0. **Thus,** \[ I_1 = \frac{296}{3} \] --- ### Step 5: Compute the second integral \[ I_2 = \int_{4}^{6} \left(44 - x^2 - \frac{4}{3}x \right) dx \] Antiderivative: \[ 44x - \frac{x^3}{3} - \frac{2}{3}x^2 \] Evaluate at \( x=6 \): \[ 44 \times 6 - \frac{216}{3} - \frac{2}{3} \times 36 \] \[ = 264 - 72 - 24 = 168 \] At \( x=4 \): \[ 44 \times 4 - \frac{64}{3} - \frac{2}{3} \times 16 \] \[ = 176 - \frac{64}{3} - \frac{32}{3} = 176 - \frac{96}{3} = 176 - 32 = 144 \] Subtract: \[ I_2=168 - 144=24 \] --- ### **Step 6: Total area** \[ A = I_1 + I_2 = \frac{296}{3} + 24 \] Express 24 as \(\frac{72}{3}\): \[ A= \frac{296}{3} + \frac{72}{3} = \frac{368}{3} \] --- ## **Final Answer:** \[ \boxed{\frac{368}{3}} \] This is the exact area of the shaded region.