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Mathematics 13 Online
OpenStudy (anonymous):

trapezoidal rule n=6 integral upper integral pi lower 0 {cos(x/2) dx

OpenStudy (anonymous):

\[\int\limits_{0}^{\pi}\cos (x/2) dx\]

OpenStudy (yuki):

if you want to use the trapezoidal rule, all you are trying to do is to estimate the area under the curve cos(x/2) using 6 trapezoids.

OpenStudy (yuki):

so that each height of the trapezoids are pi/6

OpenStudy (yuki):

since the area of a trapezoid can be calculated by the formula 1/2(b_1+b_2)h, all you have to do now is to find out b_1 and b_2 in each trapezoid.

OpenStudy (yuki):

for the left most trapezoid, b_1 will be the left base, which can be calculated by cos(0/2) = 1. b_2 would be the right base an similarly can be calculated by cos(pi/6 /2) = not a nice number. the height of the trapezoid was h = pi/6, so the area is 1/2 ( 1 + cos(pi/12))*pi/6.

OpenStudy (yuki):

let's do one more step

OpenStudy (yuki):

the next trapezoid will have b_1 which is cos(pi/12). As you noticed, it is the same as the b_2 for the first trapezoid. They have to be the same because they share that side.

OpenStudy (yuki):

Anyway, b_2 for the second trapezoid would be calculated by cos( 2*pi/6 /2) because now the x value you are plugging in is 2*pi/6.

OpenStudy (yuki):

the height is the usual, pi/6 so the area of the second trapezoid would be 1/2 (cos(pi/12) + cos(pi/6)) * pi/6

OpenStudy (yuki):

you keep doing this until you get the 6 trapezoids :)

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

to do simpsons rule same problem it will be aprox the same answer correct?

OpenStudy (yuki):

yes, but instead of a trapezoid, now you will be using a quadratic. good luck :)

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