To integrate $7y^2 \, dA$ over the triangular region $D$ with vertices at $(0,1)$, $(1,2)$, and $(4,1)$, we need to establish the bounds of integration for $y$ and $x$ based on the geometry of $D$.

Step 1: Determine Equations of the Triangle's Edges

1. Line between $(0,1)$ and $(1,2)$:

   • Slope $= \frac{2 - 1}{1 - 0} = 1$
   • Equation: $y = x + 1$

2. Line between $(1,2)$ and $(4,1)$:

   • Slope $= \frac{1 - 2}{4 - 1} = -\frac{1}{3}$
   • Equation: $y = -\frac{1}{3}x + \frac{7}{3}$

3. Line between $(0,1)$ and $(4,1)$:

   • This is a horizontal line where $y = 1$.

Step 2: Set up the Bounds

For this region, $x$ ranges from 0 to 4. For each fixed $x$, $y$ ranges from the lower edge (the line $y = 1$) to the upper edge of the triangle:

• Lower bound: $y = 1$
• Upper bound:
  • When $0 \leq x \leq 1$, $y = x + 1$
  • When $1 \leq x \leq 4$, $y = -\frac{1}{3}x + \frac{7}{3}$

Step 3: Set Up the Integral

The integral of $7y^2 \, dA$ over $D$ can be expressed as: $\int_{0}^{1} \int_{1}^{x+1} 7y^2 \, dy \, dx + \int_{1}^{4} \int_{1}^{-\frac{1}{3}x + \frac{7}{3}} 7y^2 \, dy \, dx$

Step 4: Solve the Integral

Part 1: Integral over $0 \leq x \leq 1$

$\int_{0}^{1} \int_{1}^{x+1} 7y^2 \, dy \, dx$

1. Integrate with respect to $y$: $\int 7y^2 \, dy = \frac{7y^3}{3}$
2. Apply bounds $y = 1$ to $y = x + 1$.

Part 2: Integral over $1 \leq x \leq 4$

$\int_{1}^{4} \int_{1}^{-\frac{1}{3}x + \frac{7}{3}} 7y^2 \, dy \, dx$

1. Integrate with respect to $y$: $\int 7y^2 \, dy = \frac{7y^3}{3}$
2. Apply bounds $y = 1$ to $y = -\frac{1}{3}x + \frac{7}{3}$.

Combining these, compute each integral and add the results for the final answer. Would you like a detailed evaluation of these integrals?

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1. How do we choose integration bounds based on triangular regions?
2. What changes if one of the vertices shifts?
3. How do integrals handle different slopes in region boundaries?
4. How to verify the region's area matches integral results?
5. What is the effect of increasing $y^2$ coefficient in integrand?

Tip: Drawing regions and boundaries can help visualize integration bounds more effectively.