Show that \(\int_5^1 {f(x){\rm{d}}x = - 4} \) .

Find the value of  \(\int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x} \) .

evidence of factorising 3/division by 3     A1

e.g. \(\int_1^5 {3f(x){\rm{d}}x = 3\int_1^5 {f(x){\rm{d}}x} } \) , \(\frac{{12}}{3}\) , \(\int_1^5 {\frac{{3f(x){\rm{d}}x}}{3}} \) (do not accept 4 as this is show that)

evidence of stating that reversing the limits changes the sign     A1

e.g. \(\int_5^1 {f(x){\rm{d}}x = } - \int_1^5 {f(x){\rm{d}}x} \)

\(\int_5^1 {f(x){\rm{d}}x = } - 4\)     AG     N0

[2 marks]

evidence of correctly combining the integrals (seen anywhere)     (A1)

e.g. \(I = \int_1^2 {(x + f(x)){\rm{d}}x + } \int_2^5 {(x + f(x)){\rm{d}}x = } \int_1^5 {(x + f(x)){\rm{d}}x} \)

evidence of correctly splitting the integrals (seen anywhere)     (A1)

e.g. \(I = \int_1^5 {x{\rm{d}}x + } \int_1^5 {f(x){\rm{d}}x} \) 

\(\int {x{\rm{d}}x = } \frac{{{x^2}}}{2}\) (seen anywhere)     A1

\(\int_1^5 {x{\rm{d}}x = } \left[ {\frac{{{x^2}}}{2}} \right]_1^5 = \frac{{25}}{2} - \frac{1}{2}\) \(\left( { = \frac{{24}}{2},12} \right)\)     A1

\(I =16\)     A1     N3

[5 marks]

This question was very poorly done. Very few candidates provided proper justification for part (a), a common error being to write \(\int_1^5 {f(x){\rm{d}}x = } f(5) - f(1)\) . What was being looked for was that \(\int_1^5 {3f(x){\rm{d}}x = } 3\int_1^5 {f(x){\rm{d}}x} \) and \(\int_5^1 {f(x){\rm{d}}x = } - \int_1^5 {f(x){\rm{d}}x} \) .

Part (b) had similar problems with neither the combining of limits nor the splitting of integrals being done very often. A common error was to treat \(f(x)\) as 1 in order to make \(\int_1^5 {f(x){\rm{d}}x = 4} \) and then write  \(\int_1^5 {(x + f(x)){\rm{d}}x = \left[ {x + 1} \right]} _1^5\) .