Mathematics Question and Solution
Sir Mike Ambrose is the author of the question.
y = 2^(x)
P coordinate is;
P(1, 2)
dy/dx = 2^(x)In2
For x = 1
dy/dx = 2In2
It implies;
Q coordinate is;
Q(0, (2-2In2))
Therefore;
Shaded area exactly in square units is;
(Area under the curve at x = 0 and x = 1) - (Area trapezoid with parallel sides (2-2In2) unit and 2 unit, and height 1 unit).
= (1/(In2)) - (½*1*(2+2-2In2))
= (1/(In2)) - (2-In2)
= (In2-1)²/In2 square units.
