Let's break down the problem step by step: ### **Given:** Evaluate the integral \[ \iint_D (x+y)\, dA \] where \( D \) is the region in the first quadrant bounded by: - the y-axis (\(x=0\)), - \(y = 2 - x\), - \(y = x^2.\) --- ### **Step 1: Sketch and Find Intersection Points** Set \( y = 2 - x \) equal to \( y = x^2 \): \[ x^2 = 2 - x \implies x^2 + x - 2 = 0 \implies (x+2)(x-1)=0 \] So, \( x = -2 \) or \( x = 1 \). But since we are in the first quadrant (\( x \geq 0 \)), use \( x = 0 \) (y-axis) and \( x = 1 \) (intersection). - At \( x = 0 \): \( y = 2 \) and \( y = 0 \) - At \( x = 1 \): \( y = 1 \) for both curves So, the region \( D \) is bounded: - Left: \( x=0 \) - Right: \( x=1 \) - Bottom: \( y = x^2 \) - Top: \( y = 2-x \) --- ### **Step 2: Set Up the Integral** \[ \iint_D (x+y)\, dA = \int_{x=0}^{1} \int_{y=x^2}^{y=2-x} (x + y)\, dy\, dx \] --- ### **Step 3: Evaluate the Inner Integral** \[ \int_{y=x^2}^{y=2-x} (x + y)\, dy = \left[ xy + \frac{y^2}{2} \right]_{y = x^2}^{y = 2-x} \] \[ = \left[ x(2-x) + \frac{(2-x)^2}{2} \right] - \left[ x(x^2) + \frac{(x^2)^2}{2} \right] \] Simplify: - \( x(2-x) = 2x - x^2 \) - \( (2-x)^2 = 4 - 4x + x^2 \) - \( \frac{(2-x)^2}{2} = 2 - 2x + \frac{x^2}{2} \) - \( x(x^2) = x^3 \) - \( \frac{(x^2)^2}{2} = \frac{x^4}{2} \) So, the integrand becomes: \[ [2x - x^2 + 2 - 2x + \frac{x^2}{2}] - [x^3 + \frac{x^4}{2}] \] \[ = (2x - x^2 - 2x + 2 + \frac{x^2}{2}) - x^3 - \frac{x^4}{2} \] \[ = (2 - x^2 + \frac{x^2}{2}) - x^3 - \frac{x^4}{2} \] \[ = 2 - \frac{x^2}{2} - x^3 - \frac{x^4}{2} \] --- ### **Step 4: Integrate with Respect to \( x \) from 0 to 1** \[ \int_{0}^{1} \left( 2 - \frac{x^2}{2} - x^3 - \frac{x^4}{2} \right) dx \] Integrating term by term: - \( \int 2 dx = 2x \) - \( \int \frac{x^2}{2} dx = \frac{1}{2}\cdot \frac{x^3}{3} = \frac{x^3}{6} \) - \( \int x^3 dx = \frac{x^4}{4} \) - \( \int \frac{x^4}{2} dx = \frac{1}{2}\cdot \frac{x^5}{5} = \frac{x^5}{10} \) So, \[ \left[ 2x - \frac{x^3}{6} - \frac{x^4}{4} - \frac{x^5}{10} \right]_{0}^{1} \] Plug in \( x=1 \): \[ = 2(1) - \frac{1^3}{6} - \frac{1^4}{4} - \frac{1^5}{10} = 2 - \frac{1}{6} - \frac{1}{4} - \frac{1}{10} \] Find a common denominator (60): \[ 2 = \frac{120}{60} \] \[ \frac{1}{6} = \frac{10}{60} \] \[ \frac{1}{4} = \frac{15}{60} \] \[ \frac{1}{10} = \frac{6}{60} \] So, \[ = \frac{120}{60} - \frac{10}{60} - \frac{15}{60} - \frac{6}{60} = \frac{120 - 10 - 15 - 6}{60} = \frac{89}{60} \] --- ### **Final Answer** \[ \boxed{\frac{89}{60}} \] --- #### **Graph of the Region D** - The region is bounded between \( x=0 \) and \( x=1 \). - The lower boundary is \( y = x^2 \). - The upper boundary is \( y = 2 - x \). You can sketch this by plotting \( y = x^2 \) (a parabola opening upwards) and \( y = 2 - x \) (a straight line decreasing) and shading the area between them from \( x=0 \) to \( x=1 \). --- If you want a drawn graph, let me know!