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  $\int_1^5 {(x + f(x)){\rm{d}}x = \left[ {x + 1} \right]} _1^5$ .