Let's break down the problem and solve it step-by-step.

Part (a): Find the endpoints

We are dividing the interval $[2, 4]$ into 8 equal subintervals.

The formula to calculate the width of each subinterval $\Delta x$ is: $\Delta x = \frac{b - a}{n} = \frac{4 - 2}{8} = \frac{2}{8} = 0.25$

Now, the endpoints $x_0, x_1, \dots, x_8$ will be generated by starting from 2 and adding $\Delta x = 0.25$ repeatedly. The endpoints are: $x_0 = 2, \quad x_1 = 2 + 0.25 = 2.25, \quad x_2 = 2.5, \quad x_3 = 2.75, \quad x_4 = 3, \quad x_5 = 3.25, \quad x_6 = 3.5, \quad x_7 = 3.75, \quad x_8 = 4$

Thus, the endpoints are: [ [x_0, x_1, \dots, x_8] = [2, 2.25, 2.5, 2.75, 3, 3.25, 3.5, 3.75, 4] ]

Part (b): Compute the terms $t_i = f(x_{i-1}) \Delta x$

We are using the left endpoints to compute the Riemann sum, so for each subinterval $[x_{i-1}, x_i]$, we evaluate the function at $x_{i-1}$.

The function is $f(x) = x^2 - 1$, and we have $\Delta x = 0.25$.

Now, we calculate each term $t_i = f(x_{i-1}) \Delta x$:

• $t_1 = f(x_0) \Delta x = f(2) \cdot 0.25 = (2^2 - 1) \cdot 0.25 = (4 - 1) \cdot 0.25 = 3 \cdot 0.25 = 0.75$
• $t_2 = f(x_1) \Delta x = f(2.25) \cdot 0.25 = (2.25^2 - 1) \cdot 0.25 = (5.0625 - 1) \cdot 0.25 = 4.0625 \cdot 0.25 = 1.015625$
• $t_3 = f(x_2) \Delta x = f(2.5) \cdot 0.25 = (2.5^2 - 1) \cdot 0.25 = (6.25 - 1) \cdot 0.25 = 5.25 \cdot 0.25 = 1.3125$
• $t_4 = f(x_3) \Delta x = f(2.75) \cdot 0.25 = (2.75^2 - 1) \cdot 0.25 = (7.5625 - 1) \cdot 0.25 = 6.5625 \cdot 0.25 = 1.640625$
• $t_5 = f(x_4) \Delta x = f(3) \cdot 0.25 = (3^2 - 1) \cdot 0.25 = (9 - 1) \cdot 0.25 = 8 \cdot 0.25 = 2$
• $t_6 = f(x_5) \Delta x = f(3.25) \cdot 0.25 = (3.25^2 - 1) \cdot 0.25 = (10.5625 - 1) \cdot 0.25 = 9.5625 \cdot 0.25 = 2.390625$
• $t_7 = f(x_6) \Delta x = f(3.5) \cdot 0.25 = (3.5^2 - 1) \cdot 0.25 = (12.25 - 1) \cdot 0.25 = 11.25 \cdot 0.25 = 2.8125$
• $t_8 = f(x_7) \Delta x = f(3.75) \cdot 0.25 = (3.75^2 - 1) \cdot 0.25 = (14.0625 - 1) \cdot 0.25 = 13.0625 \cdot 0.25 = 3.265625$

Thus, the terms of the Riemann sum are: [ [t_1, t_2, \dots, t_8] = [0.75, 1.015625, 1.3125, 1.640625, 2, 2.390625, 2.8125, 3.265625] ]

Part (c): Compute the value of the Riemann sum $L_8$

To get the value of the Riemann sum, sum up all the terms: $L_8 = t_1 + t_2 + t_3 + t_4 + t_5 + t_6 + t_7 + t_8$ $L_8 = 0.75 + 1.015625 + 1.3125 + 1.640625 + 2 + 2.390625 + 2.8125 + 3.265625$ $L_8 = 15.1875$

Thus, the value of the Riemann sum $L_8$ is: $L_8 = 15.19 \quad \text{(rounded to two decimal places)}$

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Summary of Answers:

• (a) Endpoints: $[2, 2.25, 2.5, 2.75, 3, 3.25, 3.5, 3.75, 4]$
• (b) Riemann sum terms: $[0.75, 1.015625, 1.3125, 1.640625, 2, 2.390625, 2.8125, 3.265625]$
• (c) Riemann sum value: $15.19$

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Would you like a detailed breakdown of any of the steps or further explanation?

Related questions:

1. How would the sum change if we used right endpoints instead of left?
2. How do you compute a Riemann sum for a different number of subintervals, such as $n=10$?
3. How can we estimate the integral using midpoint Riemann sums instead of left endpoints?
4. What is the geometric interpretation of using left-endpoint Riemann sums?
5. How does the Riemann sum compare to the exact value of the integral?

Tip:

When using Riemann sums to approximate integrals, increasing the number of subintervals generally improves the accuracy of the approximation.