Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
P(2, (16√(5)/5))
Gradient of the curve at the tangent is;
= 24/5√(5)
Therefore;
Gradient of the curve at the normal is;
= -5√(5)/24
It implies;
Coordinate Q is;
Q(4, (167√(5)/60))
Therefore area R as a single fraction in square units is;
(Area under the curve at x = 2 and x = 4) - (Area trapezium with parallel sides 167√(5)/60 unit and 16√(5)/5 unit, and height 2 unit).
= (24-⅓(8√(5))) - ½*2((167√(5)/60 )+(16√(5)/5))
= 24 - ⅓(8√(5)) - (359√(5)/60)
= (1440-519√(5))/60 square units.
