Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
Calculating k.
3k² = 108
k = 6 units.
The equation of the curve is;
y = 4x-⅓(x²)
Area R is;
Area under the curve at x=0 and x=(6-3√(2)).
= 2(6-3√(2))²-(6-3√(2))²/9
= 6(8-5√(2)) square units.
Area S is;
Area rectangle with length 12 units and width 6 units - Area trapezoid with parallel sides 12 units and 9 units, and height 3 units - Area under the curve at x=3 and x=6.
= (12*6) - ½*3(12+9) - 33
= 72 - ½(63) - 33
= 39 - ½(63)
= ½(15) square units.
It implies;
Area R ÷ Area S exactly is;
6(8-5√(2)) ÷ ½(15)
= 12(8-5√(2))/15
