(i)      Find the range of values of \(n\) for which \(\int_1^\infty  {{x^n}{\rm{d}}x} \) exists.

(ii)     Write down the value of \(\int_1^\infty  {{x^n}{\rm{d}}x} \) in terms of \(n\) , when it does exist.

Find the solution to the differential equation

\((\cos x - \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\) ,

given that \(y = - 1\) when \(x = \frac{\pi }{2}\) .

(i)     \(\int_1^b {{x^n}{\rm{d}}x}  = \left[ {\frac{{{x^{n + 1}}}}{{n + 1}}} \right]_1^b\) , \(n \ne - 1\)     M1

\( = \frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}\)     A1

\(\int_1^b {{x^n}{\rm{d}}x}  = \left[ {\ln x} \right]_1^b = \ln b\) when \(n = - 1\)     A1

if \(n + 1 > 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}} \right]\) does not exist since \({b^{n + 1}}\) increases without limit     R1

if \(n + 1 < 0,\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{b^{n + 1}}}}{{n + 1}} - \frac{1}{{n + 1}}} \right]\) exists since \({b^{n + 1}} \to 0\) as \(b \to \infty \)     R1

if \(n =  - 1\) , \(\mathop {\lim }\limits_{b \to \infty } \left[ {\ln b} \right]\) does not exist since \({\ln b}\) increases without limit     R1

(so integral exists when \(n < - 1\) )

(ii)     \(\int_1^b {{x^n}{\rm{d}}x}  = \frac{1}{{n + 1}}\) , \((n < - 1)\)     A1

[7 marks]

\((\cos x - \sin x)\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + (\cos x + \sin x)y = \cos x + \sin x\)

\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \frac{{\cos x + \sin x}}{{\cos x - \sin x}}y = \frac{{\cos x + \sin x}}{{\cos x - \sin x}}\)     M1

IF \({{\rm{e}}^{\int {\frac{{\cos x + \sin x}}{{\cos x - \sin x}}} {\rm{d}}x}} = {{\rm{e}}^{ - \ln (\cos x - \sin x)}} = \frac{1}{{\cos x - \sin x}}\)     M1A1A1

\(\frac{y}{{\cos x - \sin x}} = \int {\frac{{\cos x + \sin x}}{{{{(\cos x - \sin x)}^2}}}} {\rm{d}}x\)     (M1)

\(\frac{1}{{\cos x - \sin x}} + k\)     A1

Note: Award the above A1 even if \(k\) is missing.

\(y = 1 + k(\cos x - \sin x)\)

\(x = \frac{\pi }{2}\) , \(y =  - 1\)

\( - 1 = 1 + k( - 1)\)     M1

\(k = 2\)

\(y = 1 + 2(\cos x - \sin x)\)     A1

Note: It is acceptable to solve the equation using separation of variables.

[8 marks]

This was found to be the most difficult question on the paper. Whilst the question looked straightforward, indiscriminate use of the infinity symbol showed a lack of appreciation of the subtleties involved in the question. Many of the refinements required in each section were not considered. Knowledge of improper integrals was very poor. Perhaps integration is seen too easily as the application of a number of rules without much thought.

Generally this was quite well done. However, many candidates did not realize that it could be solved using an integrating factor and used substitution or variables separable. While these last two approaches could work, the algebra involved soon became unmanageable. This led to many mistakes.