The goal of this exercise is to approximate the value of the definite integral  ∫42(x2−1)dx  using a Riemann sum with left endpoints and  8  subintervals (i.e. using the Riemann sum  L8 ).

a) If you sub-divide the interval  [2,4]  into  8  subintervals of equal length  Δx , you get  8  subintervals of the form  [xi−1,xi]  whose union is the interval [2,4] .  Find the endpoints  x0,x1,…,x8  that define these  8  subintervals.
FORMATTING: To enter your answer, list the endpoints in the form  [x0,x1,…,x8]  including the square brackets [ ] and a comma (,) between endpoints.  The endpoints should be exact numbers, ordered from left to right.
Answer:     [x0,x1,…,x8]=    Preview

b) Compute the exact value of each term  ti=f(xi−1)Δx  of the Riemann sum  L8 .
FORMATTING: To enter your answer, list the terms in the form  [t1,t2,…,t8]  including the square brackets [ ] and a comma (,) between consecutive terms.  Each term must be an exact number.
Answer:     [t1,t2,…,t8]=    Preview

c) Compute the value of the Riemann sum  L8 . Write your answer with an accuracy of two decimal places.
Answer:

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.

The goal of this exercise is to approximate the value of the definite integral ∫42(x^2−1)dx using a Riemann sum with left endpoints and 8 subintervals (i.e. using the Riemann sum L8).

(a) If you sub-divide the interval [2,4] into 8 subintervals of equal length Δx, you get 8 subintervals of the form [xi−1,xi] whose union is the interval [2,4]. Find the endpoints x0,x1,…,x8 that define these 8 subintervals.

(b) Compute the exact value of each term ti=f(xi−1)Δx of the Riemann sum L8.

(c) Compute the value of the Riemann sum L8. Write your answer with an accuracy of two decimal places.

Learn how to approximate the definite integral ∫(x^2−1)dx from 2 to 4 using a Riemann sum with 8 subintervals and left endpoints. This solution breaks down the steps, including calculating subinterval lengths, evaluating each term in the sum, and computing the total Riemann sum, ideal for Calculus I students.

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