To solve this problem using a lower sum with two rectangles of equal width, let's proceed with the following steps:

1. Determine the interval width: We are given the interval $[0, 4]$ and want two rectangles, so each rectangle will cover a width of: $\Delta x = \frac{4 - 0}{2} = 2$

2. Calculate the left endpoints for the lower sum: Since this is a lower sum, we use the left endpoint of each interval to find the height of each rectangle.

   • The first rectangle covers the interval $[0, 2]$ with left endpoint $x = 0$.
   • The second rectangle covers the interval $[2, 4]$ with left endpoint $x = 2$.

3. Evaluate $f(x) = 7x^2$ at each left endpoint:

   • For $x = 0$: $f(0) = 7 \cdot 0^2 = 0$.
   • For $x = 2$: $f(2) = 7 \cdot 2^2 = 7 \cdot 4 = 28$.

4. Calculate the area of each rectangle and sum them up:

   • Area of the first rectangle: $f(0) \cdot \Delta x = 0 \cdot 2 = 0$.
   • Area of the second rectangle: $f(2) \cdot \Delta x = 28 \cdot 2 = 56$.

5. Sum of the areas for the lower sum: $\text{Total Area} = 0 + 56 = 56$

Thus, the estimated area using a lower sum with two rectangles of equal width is 56 square units.

Would you like a breakdown of each step or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would the result change if we used an upper sum instead of a lower sum?
2. What is the impact of increasing the number of rectangles on the accuracy of the approximation?
3. Can you derive the area using a lower sum with four rectangles of equal width?
4. What would be the area estimate if we used the midpoint rule instead of the left or right endpoints?
5. How does this approach relate to finding the definite integral of $f(x) = 7x^2$ from 0 to 4?

Tip: Using more rectangles generally provides a more accurate approximation of the area under the curve, as it reduces the error between the rectangles and the actual curve.