(a) Use the Midpoint Rule and the given data to estimate the value of the integral I. (Give the answer to two decimal places.)

I = int_0^3.2 f(x)dx
1
x f(x) x f(x)
0.0 7.3 2.0 8.8
0.4 6.8 2.4 7.3
0.8 8.1 2.8 7.6
1.2 7.6 3.2 7.3
1.6 8.2
(b) If it is known that -5 ≤ f''(x) ≤ 3 for all x, estimate the error involved in the approximation in part (a). (Give the answer to four decimal places.)
2

(a) 24.45 (b) 0.0045

To estimate the value of the integral I using the Midpoint Rule, we need to divide the interval [0, 3.2] into subintervals of equal width and use the midpoints of these subintervals to evaluate the function.

Step 1: Determine the number of subintervals.
Since the interval goes from 0 to 3.2, and we have data points at 0.0, 0.4, 0.8, 1.2, 1.6, 2.0, 2.4, 2.8, and 3.2, we can use 9 subintervals.

Step 2: Calculate the width of each subinterval.
The width of each subinterval is given by (b - a) / n, where b is the upper limit (3.2), a is the lower limit (0.0), and n is the number of subintervals (9).
So, the width of each subinterval is (3.2 - 0.0) / 9 = 0.3556 (rounded to four decimal places).

Step 3: Find the midpoints of each subinterval.
The midpoints can be found by adding half of the width to the starting point of each subinterval. Therefore, the midpoints are:
0.1778, 0.5333, 0.8889, 1.2444, 1.6000, 1.9556, 2.3111, 2.6667, 3.0222.

Step 4: Evaluate the function at the midpoints.
Using the given data, we can approximate the values of the function at each midpoint:
f(0.1778) = 7.3
f(0.5333) = 6.8
f(0.8889) = 8.1
f(1.2444) = 7.6
f(1.6000) = 8.2
f(1.9556) = missing
f(2.3111) = 8.8
f(2.6667) = 7.3
f(3.0222) = 7.6

Step 5: Calculate the approximate value of the integral using the Midpoint Rule.
The Midpoint Rule formula for approximating an integral is:
I ≈ Δx * (f(x1) + f(x2) + ... + f(xn)),
where Δx is the width of each subinterval and f(xi) is the value of the function at the ith midpoint.

Substituting the values into the formula, we have:
I ≈ 0.3556 * (7.3 + 6.8 + 8.1 + 7.6 + 8.2 + 2 * 8.8 + 7.3 + 7.6) = 18.43 (rounded to two decimal places).

Therefore, the estimated value of the integral I is 18.43.

Now, let's move on to calculating the error involved in the approximation.

(b) To estimate the error in the approximation made in part (a), we can use the formula for the maximum error bound when applying the Midpoint Rule.

The maximum error bound is given by the formula:
E ≤ (K * Δx^2 * (b - a))/ 24,
where K is the maximum value of the second derivative of the function within the interval [a, b], Δx is the width of each subinterval, and (b - a) is the overall width of the interval.

Since it is given that -5 ≤ f''(x) ≤ 3 for all x, we can take K = 3 as the maximum value of the second derivative within the given interval.

Using the values Δx = 0.3556 (from part (a)), a = 0, and b = 3.2, we can calculate the error:

E ≤ (3 * 0.3556^2 * (3.2 - 0)) / 24 = 0.0106 (rounded to four decimal places).

Therefore, the estimated error involved in the approximation from part (a) is 0.0106.

(a) To estimate the value of the integral I using the Midpoint Rule, we can divide the interval [0, 3.2] into subintervals of equal width and approximate the area under the curve f(x) using the midpoint of each subinterval.

First, we need to calculate the width of each subinterval:
Δx = (b - a) / n
Δx = (3.2 - 0) / 8
Δx = 0.4

Next, we can calculate the midpoint of each subinterval:
x_1 = 0.2
x_2 = 0.6
x_3 = 1.0
x_4 = 1.4
x_5 = 1.8
x_6 = 2.2
x_7 = 2.6
x_8 = 3.0

Now, we can calculate the approximate value of the integral using the Midpoint Rule formula:
I ≈ Δx * (f(x_1) + f(x_2) + f(x_3) + f(x_4) + f(x_5) + f(x_6) + f(x_7) + f(x_8))

I ≈ 0.4 * (f(0.2) + f(0.6) + f(1.0) + f(1.4) + f(1.8) + f(2.2) + f(2.6) + f(3.0))

I ≈ 0.4 * (6.8 + 8.1 + 7.6 + 8.2 + 7.3 + 7.6 + 7.3 + 7.3)

I ≈ 0.4 * 60.2

I ≈ 24.08

Therefore, the estimated value of the integral I is approximately 24.08.

(b) To estimate the error involved in the approximation from part (a), we can use the error bound formula for the Midpoint Rule:

Error ≤ (K * Δx^3) / 24

Where K is the maximum value of |f''(x)| over the interval [0, 3.2], and Δx is the width of each subinterval.

Given that -5 ≤ f''(x) ≤ 3 for all x, the maximum value of |f''(x)| is 5.

Plugging in the values, we have:
Error ≤ (5 * 0.4^3) / 24

Error ≤ 0.04 / 24

Error ≤ 0.0016667

Therefore, the estimated error involved in the approximation from part (a) is approximately 0.0017.