Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
P(2, (8/3))
Q(4, (32/15))
Gradient of the curve is;
dy/dx = -4x/(x²-1)²
For x = 2.
dy/dx = -8/9
It implies;
R(5, 0)
Area S exactly in square units is;
Area under the curve at x = 2 and x = 4 - Area trapezoid with parallel sides (8/3) units and (8/9) unit, and height 2 units.
= 4 + ln(9/5) - ½*2((8/3)+(8/9))
= 4 + ln(9/5) - (32/9)
= (4/9) + ln(9/5)
= (4+9(ln(9/5)))/9 square units.
