Given the revised values:

• $f(-7) = 4$
• $f(-4) = 4$
• $f(1) = 2$
• $f(3) = 1$

We can now recalculate the left-end approximation $L_n$.

Step 1: Recalculate the Left Riemann Sum $L_n$

Using the partition points $\{-7, -4, 1, 3, 7\}$ and the given function values:

$L_n = f(-7) \cdot \Delta x_1 + f(-4) \cdot \Delta x_2 + f(1) \cdot \Delta x_3 + f(3) \cdot \Delta x_4$

Substituting the function values and the widths of the subintervals:

• $\Delta x_1 = 3$
• $\Delta x_2 = 5$
• $\Delta x_3 = 2$
• $\Delta x_4 = 4$

Step 2: Substitute the Values into the Formula

$L_n = (4) \cdot 3 + (4) \cdot 5 + (2) \cdot 2 + (1) \cdot 4$

Step 3: Perform the Calculations

$L_n = 12 + 20 + 4 + 4 = 40$

Final Answer

The left-end approximation $L_n$ of the area under the curve is 40.

Would you like more details or have any other questions?

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Related Questions:

1. How does the Riemann sum approximation change with different partitions?
2. If you had a finer partition, how would $L_n$ be affected?
3. How can you verify the accuracy of this approximation?
4. How does the shape of $f(x)$ affect the Riemann sum approximation?
5. What are the advantages of using a left Riemann sum over other types?
6. How would you graphically interpret the sum $L_n = 40$?
7. Can the Riemann sum be applied to functions with more complex shapes?
8. How would $L_n$ change if the function values were negative?

Tip: Always check the consistency of your partition widths and function values before calculating the Riemann sum to avoid errors in approximation.