Using the midpoint rule?

Question

se the midpoint rule with the given value of n to approximate the integral

interval from 0 to Pie(3.14) sec(x/3) dx n+6

how would you work this problem.

Answer 1

∫ sec(x/3) dx , limits [0,PI]

divide the interval [0,PI] into 6 sub-intervals

(PI-0)/6

[0,PI/6],[PI/6,PI/3],{PI/3,PI/2],[PI/2...

Take the midpoint of each sub-interval.

PI/12, PI/4, 5PI/12, 7PI/12, 9PI/12, 11PI/12

Evaluate f(x)=sec(x/3) for each of the above.

f(PI/36) = 1.0038198375433473

f(PI/12) =- 1.035276180410083

p(5PI/36) = 1.2150906493338438

f(7PI/36) = 1.220774588761456

f(9PI/36) = 2.1518186743350464

f(11PI/36) = 1.7434467956210977

(PI-0)/2[f(PI/36)+f(PI/12)+f(5PI/36)+f...

+f(9PI/16+f(11PI/36)]

add all the f's and multiply by PI/2

8.370226726004874*(PI/6)

=4.38264046521631

The formula is correct. Please check all of my calculations.