 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/20/a/solution

Let

a = π /3

b = 2 π /3

and take

n

equal subintervals of width

h = (b - a) / n = π / (3 n)

. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sums

J = lim_{n \to \infty} h \sum_{j = 0}^{n - 1} sin (a + jh) .

Using Euler's formula, the corresponding complex sum is a finite geometric series:

S_{n} = h \sum_{j = 0}^{n - 1} e^{i (a + jh)} = h e^{ia} \frac{1 - e ^{inh}}{1 - e ^{ih}}, nh = b - a = \frac{π}{3} .

Since

(e^{ih} - 1) / h \to i

lim S_{n} = \frac{e ^{ib} - e ^{ia}}{i} = \frac{( - 1/2 + i 3 /2 ) - ( 1/2 + i 3 /2 )}{i} = i .

J = 1.

One can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum gives

h \sum_{j = 0}^{n - 1} sin (a + jh) = h \frac{s i n ( nh /2 )}{s i n ( h /2 )} sin (a + \frac{n - 1}{2} h) = \frac{h}{2} cot \frac{h}{2} ⟶ 1.

Thus the evaluation is explicitly by a Riemann sum, as required.

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