The graphs of $f(x) = - {x^2} + 2$ and $g(x) = {x^3} - {x^2} - bx + 2,{\text{ }}b > 0$, intersect and create two closed regions. Show that these two regions have equal areas.

to find the points of intersection of the two curves

$- {x^2} + 2 = {x^3} - {x^2} - bx + 2$     M1

${x^3} - bx = x({x^2} - b) = 0$

$\Rightarrow x = 0;{\text{ }}x = \pm \sqrt b$     A1A1

${A_1} = \int_{ - \sqrt b }^0 {\left[ {({x^3} - {x^2} - bx + 2) - ( - {x^2} + 2)} \right]} {\text{d}}x\left( { = \int_{ - \sqrt b }^0 {({x^3} - bx){\text{d}}x} } \right)$     M1

$= \left[ {\frac{{{x^4}}}{4} - \frac{{b{x^2}}}{2}} \right]_{ - \sqrt b }^0$

$= - \left( {\frac{{{{( - \sqrt b )}^4}}}{4} - \frac{{b{{( - \sqrt b )}^2}}}{2}} \right) = - \frac{{{b^2}}}{4} + \frac{{{b^2}}}{2} = \frac{{{b^2}}}{4}$     A1

${A_2} = \int_0^{\sqrt b } {\left[ {( - {x^2} + 2) - ({x^3} - {x^2} - bx + 2)} \right]{\text{d}}x}$     M1

$= \int_0^{\sqrt b } {( - {x^3} + bx){\text{d}}x}$

$= \left[ { - \frac{{{x^4}}}{4} + \frac{{b{x^2}}}{2}} \right]_0^{\sqrt b } = \frac{{{b^2}}}{4}$     A1

therefore ${A_1} = {A_2} = \frac{{{b^2}}}{4}$     AG

[7 marks]

Most candidates knew how to tackle this question. The most common error was in giving +b and –b as the x-coordinates of the point of intersection.