Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 7

Table of contents
- a 7
  - Solution a
- b 7
  - Solution b

a

Solution

The derivative of

cot u

is

- csc^{2} u

. The chain rule therefore gives

\int \frac{d x}{sin ^{2} ( 3 x + 1 )} = - \frac{1}{3} cot (3 x + 1) + C .

The antiderivative applies on any interval containing no zero of

sin (3 x + 1)

.

b

Solution

Additivity of the definite integral combines the adjacent intervals into

[0, n]

. For an integer

n \geq 1

, the power rule gives

\sum_{j = 0}^{n - 1} \int_{j}^{j + 1} (x - 2)^{n} d x = \int_{0}^{n} (x - 2)^{n} d x = \frac{( n - 2 ) ^{n + 1} - ( - 2 ) ^{n + 1}}{n + 1} .

For

n = 0

, interpret the indexed sum as an empty sum, equal to zero; the displayed expression also gives zero. This convention resolves the otherwise degenerate endpoint of the stated nonnegative-integer range.

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