Question

The diagram shows the curve y=x3 −6x2 +9xforx≥0. The curve has a maximum point at A and a minimum point on the x-axis at B. The normal to the curve at C (2, 2) meets the normal to the curve at B at the point D.
Find the area of shaded region.

Diagram question 11 http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s09_qp_1.pdf

Answer 1

If you factor the function---> y = x(x-3)^2
this shows zeros at x=0 and x=3
B is (3,0) and the normal at B is the line x=3

taking derivative ----> y' = 3x^2 -12x +9
set equal to 0 to find local max , A
3x^2 -12x +9 = 0
x^2 -4x +3 = 0
(x-3)(x-1) = 0
x=3 is the min, point B
x=1 is max, point A
A is (1,4)

the slope of tangent line at point C is:
y'(2) = -3
slope of normal is opp reciprocal of tangent
slope = 1/3
equation of line at point C
---> y-2 = 1/3(x-2)
--> y = 1/3x + 4/3

point D is intersection of normal lines through C and B
plug in x=3 into normal line through C
y(3) = 7/3
D is (3, 7/3)

Area can be found by subtracting area under curve (integrating) from area of combined region
(trapezoid)
\[\int\limits_2^3 x^3 -6x^2 +9x dx = |_2^3 \frac{x^4}{4} - 2x^3 +\frac{9}{2} x^2 = \frac{3}{4}\]
trapezoid can be split into rectangle (1 by 2) with small triangle (1 by 1/3)
area rectangle = 2
area triangle = 1/6
area trapezoid = 13/6

--> area region = 13/6 - 3/4 = 17/12

Answer 2

Can you pleased ketch out the rectangle and the trapezoid?

Answer 3

please sketch*