Use the trapezoidal rule, the midpoint rule, and simpson's rule to approximate the given integral with the specified value of n. (round your answers to six decimal places.) 5 4 1 ln(t) dt, n = 10 (a) the trapezoidal rule (b) the midpoint rule (c) simpson's rule

I guess the "5" is supposed to represent the integral sign?

With subintervals, we split up the domain of integration as

[1, 13/10], [13/10, 8/5], [8/5, 19/10], ... , [37/10, 4]

For each rule, it will help to have a sequence that determines the end points of each subinterval. This is easily, since they form arithmetic sequences. Left endpoints are generated according to

and right endpoints are given by

where .

a. For the trapezoidal rule, we approximate the area under the curve over each subinterval with the area of a trapezoid with "height" equal to the length of each subinterval, , and "bases" equal to the values of at both endpoints of each subinterval. The area of the trapezoid over the -th subinterval is

Then the integral is approximately

b. For the midpoint rule, we take the rectangle over each subinterval with base length equal to the length of each subinterval and height equal to the value of at the average of the subinterval's endpoints, . The area of the rectangle over the -th subinterval is then

so the integral is approximately

c. For Simpson's rule, we find a quadratic interpolation of over each subinterval given by

where is the midpoint of the -th subinterval,

Then the integral is equal to the sum of the integrals of each interpolation over the corresponding -th subinterval.

It's easy to show that

so that the value of the overall integral is approximately

no problobm

step-by-step explanation:

idk

step-by-step explanation: