Core 2 Bemusement

Hi! Okay, so I'm going through a paper - OCR MEI Core 2 January 2013:

http://www.mei.org.uk/files/papers/C2_Paper_Jan13.pdf

Q10 part ii on pdf reader page 4): WHAAT?!?! 19.3333... how? O.O

I integtated the 'normal' -(1/4)x+4 to -(1/8)x^2+4x and called it A
I integrated the 'curve' y=x^2-4x+3 to (1/3)x^3-2x^2+3x and called it B

My original limits gave the wrong answer, so I set up a spreadsheet in excel with 3 possible limits:

The letters are my names for specific numbers, e.g. C is 32

16: into A gave 32 (C), and into B gave 901.333333... (D)
4: into A gave 14 (E), and into B gave 1.3333333 (F)
3: into A gave 10.875 (G), and into B gave 0 (h)

Then I did a series of calculations:

Substituting for A and B:
limit 16 to 4:C-D-E+F=-882
limit 16 to 3:C-D-G+H=-880.2083333…
limit 4 to 3: E-F-G+H=1.791666667

Substituting for A only:
limit 16 to 4:C-E=18
limit 16 to 3:C-G=21.125
limit 4 to 3: E-G=3.125

Substituting for B only:
limit 16 to 4-F=900
limit 16 to 3-H=901.3333333…
limit 4 to 3: F-H=1.3333333…

None of these give me 19 and 1/3. Where on earth did they get the area for 9 and 1/3? And why have they only used equation B in the mark scheme – the question states that we’re looking for the region ABC bounded by the curve and the normal? Why have they used the 4 to 3 limits? I've tried everything I can think of! @.@
The coordinates are:
A(4,3)
B(3,0)
C(16, 0)
Hence the choice of limits.
This is the mark scheme (10 part ii is on pdf reader page 31):

http://www.mei.org.uk/files/papers/C2_Paper_Jan13.pdf

The mark scheme has rendered me absolutely stunned, and so has the question. Can anyone help me out, please?

I'd find the area of the triangle, find the integral of the curve from A to B, then add them together. It'd be much simpler than messing around with the integral of the normal

Goodness me - that was quick! Thank you so much for your reply - you've made it a whole lot simpler! I guess I never looked at it that way.

I broke the shape up into two figures, a right triangle with the vertices (4,0), (4,3), and (0,16) and the area under the parabola between x=x=3 and x=4. The first area can be found using the formula for area of a triangle and the second can be found using integration. When added together, the combined area does come out to be 19 1/3.