Please accept this question if you can give step by step cor

Please accept this question if you can give step by step correct solution of this question.I posted this question earlier few minutes back and got answer in poor handwriting(can even read it) and i am afraid that they are wrong as well. Please if possible dont give handwritten solution.Typed out solution would be really appreciated :)

Solution

Since, f(x)= sin(2x) is continuous on the interval [0,1/2] so it the Riemann integral exists and given partition is {0,1/12,1/8,1/4,3/8,1/2},so

Riemann Sum = (1/12-0) sin(2A) + (1/8-1/12) sin(2B) + (1/4-1/8) sin(2C) + (3/8-1/4) sin(2D) + (1/2-3/8) sin(2E)

where A,B,C,D & E are arbitrary real numbers between the intervals [0,1/12],[1/12,1/8],[1/8,1/4],[1/4,3/8] & [3/8,1/2] respectively . You can coose any number between them. This means you can arbitrary many Riemann Sums for this particular questions.

Here, I\'m giving you the maximum possible one, with A=1/12 , B=1/8, C=1/4, D=1/4 & E=3/8, which is

(1/12-0) sin(2/12) + (1/8-1/12) sin(2/8) + (1/4-1/8) sin(2/4) + (3/8-1/4) sin(2/4) + (1/2-3/8) sin(2*3/8) = (5+2)/12.

Do remember there can be more answers.