Answer:
Explanation:
a. To compute M₁ for y = f(x) on the interval [2, 5], we need to divide the interval into n = 3 subintervals of equal width Δx = (5 - 2)/3 = 1. Then, we can evaluate f(x) at the endpoints and midpoints of the subintervals to get the heights of the corresponding rectangles.
The values of Ax and the locations of the subintervals are:
Ax = Δx [f(2) + f(3) + f(4)] = 1[22 + 25 + 28] = 75
The sketch of the function and the rectangles is shown below:
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2 3 4 5
b. The area of the region bounded by y = f(x) and the x-axis on [2, 5] is given by the definite integral:
∫[2,5] f(x) dx
= ∫[2,5] (3x + 4) dx
= [3/2 x^2 + 4x] from 2 to 5
= (3/2 (5)^2 + 4(5)) - (3/2 (2)^2 + 4(2))
= 37.5
c. The values computed in (a) and (b) turn out to be the same because the function f(x) = 3x + 4 is a linear function, which means that the area under the curve between any two points is the same as the area of the corresponding trapezoid with height equal to the average of the function values at those points. This is true regardless of the number of subintervals used in the approximation.
If we use a number different than n = 4 and compute M, we will get a different value for the area approximation. However, if the function is continuous and integrable on the interval, the exact area will still be given by the definite integral.
L4 and R4 will not have the same value as the exact area of the region found in (b) because they use the left and right endpoints of each subinterval to determine the height of the corresponding rectangle, which can overestimate or underestimate the true area. M4, on the other hand, uses the midpoint of each subinterval, which tends to give a better approximation to the true area.