The function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Trapezoidal Approximation, using the 4 intervals between those given points.

x 4 9 11 14 15
f(x) -6 -11 -18 -21 -25

Steve u r so unhelpful smh

To estimate the average value of the function f using a Trapezoidal Approximation, we need to calculate the area under the curve represented by the given points on the interval [4, 15] and then divide it by the length of the interval.

First, let's calculate the widths of the intervals. The interval [4, 15] is divided into 4 intervals, so each interval has a width of (15 - 4) / 4 = 2.75.

Next, let's calculate the areas of the trapezoids for each interval and sum them up.

For the first interval [4, 6.75]:
Width = 2.75
Average height = (f(4) + f(9)) / 2 = (-6 - 11) / 2 = -17/2
Area = Width * Average height = 2.75 * (-17/2) = -46.75/2

For the second interval [6.75, 9]:
Width = 2.75
Average height = (f(9) + f(11)) / 2 = (-11 - 18) / 2 = -29/2
Area = Width * Average height = 2.75 * (-29/2) = -79.75/2

For the third interval [9, 11.75]:
Width = 2.75
Average height = (f(11) + f(14)) / 2 = (-18 - 21) / 2 = -39/2
Area = Width * Average height = 2.75 * (-39/2) = -106.75/2

For the fourth interval [11.75, 15]:
Width = 2.75
Average height = (f(14) + f(15)) / 2 = (-21 - 25) / 2 = -23/2
Area = Width * Average height = 2.75 * (-23/2) = -62.75/2

Now let's sum up the areas of the trapezoids:
Total Area = (-46.75/2) + (-79.75/2) + (-106.75/2) + (-62.75/2) = -295/2

Finally, let's calculate the average value by dividing the total area by the length of the interval:
Average Value = Total Area / Interval Length = (-295/2) / (15 - 4) = (-295/2) / 11 = -295/22 ≈ -13.409

Therefore, the estimated average value of the function f using a Trapezoidal Approximation with 4 intervals is approximately -13.409.

To estimate the average value of the function using a Trapezoidal Approximation, we need to calculate the area under the curve by approximating it with trapezoids.

First, let's calculate the width of each interval. Since we have 4 intervals between 5 points, each interval will have a width of (15 - 4) / 4 = 2.75.

Next, for each interval, we'll calculate the average value of f(x) by taking the average of the heights of the two endpoints. The heights are given as -6, -11, -18, -21, and -25.

The average value for the first interval (4 to 6.75) is (f(4) + f(6.75)) / 2 = (-6 + -11) / 2 = -8.5.
The average value for the second interval (6.75 to 9) is (f(6.75) + f(9)) / 2 = (-11 + -18) / 2 = -14.5.
The average value for the third interval (9 to 11) is (f(9) + f(11)) / 2 = (-18 + -21) / 2 = -19.5.
The average value for the fourth interval (11 to 13.75) is (f(11) + f(13.75)) / 2 = (-21 + -25) / 2 = -23.

Finally, we'll calculate the sum of these average values and multiply it by the width of each interval to get the approximate area under the curve:

Average value * interval width = (-8.5 × 2.75) + (-14.5 × 2.75) + (-19.5 × 2.75) + (-23 × 2.75)
= -23.375 + -39.875 + -53.625 + -63.25
= -180.125

Therefore, the average value of the function with this Trapezoidal Approximation is approximately -180.125.

for each trapezoid number i the area is

(f(x_i)+f(x_i+1))/2 * (x_i+1 - x_i)