1. **Problem statement:** We have the line $y = 3x + 4$ and shaded rectangles under the line from $x=0$ to $x=50$. Each rectangle's height is the function value at the left side of the rectangle, so the height at $x=k$ is $3k + 4$. We want to find the total area of all these rectangles. 2. **Formula and approach:** The total area is the sum of the areas of all rectangles. Each rectangle has width 1 and height $3k + 4$ for $k=0,1,2,\ldots,49$ (since the last rectangle starts at $x=49$ and ends at $x=50$). 3. **Sum expression:** Total area = $$\sum_{k=0}^{49} (3k + 4) \times 1 = \sum_{k=0}^{49} (3k + 4)$$ 4. **Separate the sum:** $$\sum_{k=0}^{49} (3k + 4) = 3 \sum_{k=0}^{49} k + 4 \sum_{k=0}^{49} 1$$ 5. **Calculate each sum:** - Number of terms: $50$ (from $k=0$ to $k=49$) - Sum of integers: $$\sum_{k=0}^{49} k = \frac{49 \times 50}{2} = 1225$$ - Sum of ones: $$\sum_{k=0}^{49} 1 = 50$$ 6. **Substitute values:** $$3 \times 1225 + 4 \times 50 = 3675 + 200 = 3875$$ 7. **Final answer:** The total area of the shaded rectangles is **3875**.