i will tel you HOW to do this, but you will have to actually DO it yourself.

we have an interval [0,π/2], and we are going to chop it up into 3 equal pieces.

this will becomes the THREE intervals [0,π/6], [π/6,π/3] and [π/3,π/2].

so our right-hand endpoints are:

π/6, π/3 and π/2.

we are going to approximate the area between the curve f(x) = cos(2x) and the x-axis, by three rectangles:

the rectangles will have HEIGHT f(x_{i}), where x_{i}is one of our 3 endpoints, and each rectangle will have a BASE (width) of length (1/3)(π/2) = π/6.

so we're going to have 3 terms in our sum, each one will look like:

cos(2x_{i})(π/6). i leave it to you to put this in "sigma" notation.

the GENERAL form of a riemann sum from a to b is:

where x_{1}= a, and x_{n+1}= b, and for each i, x_{i}< x_{i+1}.

often, the sub-intervals [x_{i},x_{i+1}] are all chosen to be the same size, while x_{i}* is allowed to be ANY point in [x_{i},x_{i+1}].

it is, however, customary to pick a sure-fire way of coming up with some "organized" way of picking the x_{i}* (such as left endpoint, right endpoint, midpoint, maximum value or minimum value). if f is a continuous function, and n is large enough, it won't make "too much difference" which point we pick, because all these values will be "close together".