Andy C. answered 06/08/18
Tutor
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Math/Physics Tutor
Divide the interval [3,13] into n=5 subintervals of length Δx=2, with the following endpoints: a=3,5,7,9,11,13=b.
Now, we just evaluate the function at those endpoints:
f(x0)=f(a)=f(3)=sin(3)+15=15.1411200080599
2f(x1)=2f(5)=2sin(5)+50=48.0821514506737
2f(x2)=2f(7)=2sin(7)+70=71.3139731974376
2f(x3)=2f(9)=2sin(9)+90=90.8242369704835
2f(x4)=2f(11)=2sin(11)+110=108.000019586899
f(x5)=f(b)=f(13)=sin(13)+65=65.4201670368266
Now, we just evaluate the function at those endpoints:
f(x0)=f(a)=f(3)=sin(3)+15=15.1411200080599
2f(x1)=2f(5)=2sin(5)+50=48.0821514506737
2f(x2)=2f(7)=2sin(7)+70=71.3139731974376
2f(x3)=2f(9)=2sin(9)+90=90.8242369704835
2f(x4)=2f(11)=2sin(11)+110=108.000019586899
f(x5)=f(b)=f(13)=sin(13)+65=65.4201670368266
Finally, just sum up the above values and multiply by Δx2=1: 1(15.1411200080599+48.0821514506737+71.3139731974376+90.8242369704835+108.000019586899+65.4201670368266)=398.78166825038
Answer: 398.781668250380
Answer: 398.781668250380
THe anti-derivative is (5/2)x^2 - cos(x)
(5/2)(13)^2 - cos(13) - [ (5/2)(3)^2 - cos(3) ] =
(5/2)(169) - cos(13) - (5/2)(9) + cos(3) =
398.10256072194935832887574104979
(5/2)(13)^2 - cos(13) - [ (5/2)(3)^2 - cos(3) ] =
(5/2)(169) - cos(13) - (5/2)(9) + cos(3) =
398.10256072194935832887574104979
which is an error of about 0.17% < 1/4 %