To solve the problem, we need to approximate the area under the curve using the Left Endpoint Approximation method with 5 subintervals.

Step 1: Identify the interval and subintervals

• The interval given is from $x = 2$ to $x = 7$.
• The number of subintervals is 5.

The width of each subinterval $\Delta x$ is calculated as: $\Delta x = \frac{7 - 2}{5} = \frac{5}{5} = 1$

Step 2: Determine the left endpoints

For 5 subintervals, the left endpoints will be:

• $x_1 = 2$
• $x_2 = 3$
• $x_3 = 4$
• $x_4 = 5$
• $x_5 = 6$

Step 3: Evaluate the function at each left endpoint

From the graph, approximate the value of $f(x)$ at each left endpoint:

• $f(2) = 1$
• $f(3) = 2$
• $f(4) = 3$
• $f(5) = 4$
• $f(6) = 5$

Step 4: Calculate the area of each rectangle

Each rectangle has an area given by: $\text{Area} = f(x_i) \times \Delta x$ So, the total area approximation $A$ is: $A = f(2) \times 1 + f(3) \times 1 + f(4) \times 1 + f(5) \times 1 + f(6) \times 1$ $A = 1 \times 1 + 2 \times 1 + 3 \times 1 + 4 \times 1 + 5 \times 1$ $A = 1 + 2 + 3 + 4 + 5 = 15$

Conclusion

The approximate area under the curve from $x = 2$ to $x = 7$ using a Left Endpoint Approximation with 5 subintervals is 15 square units.

Would you like further details or have any questions?

Here are some related questions:

1. What is the difference between left and right endpoint approximations?
2. How would the approximation change if we used more subintervals?
3. What is the exact area under the curve using integration?
4. How does the choice of endpoint affect the accuracy of the approximation?
5. Can you explain the trapezoidal rule and how it compares to the left endpoint approximation?
6. What happens if we use midpoint instead of the left endpoint for this approximation?
7. How would the result change if the curve was concave down?
8. Can we apply the Left Endpoint Approximation to a decreasing function?

Tip: The more subintervals you use, the closer your approximation will be to the actual area under the curve.