Equate x − 4 to √(2x).
Then gain (x − 4)2 = 2x.
Next, x2 − 8x +16 − 2x or x2 − 10x +16 = 0.
Take the roots of x2 − 10x +16 = 0 as
{-(-10) ± √[(-10)2 − 4(1)(16)]} ÷ 2(1).
Simplify to {10 ± √[36]} ÷ 2(1) or
{10 ± 6} ÷ 2 which gives 8 and 2
as roots of x2 − 10x +16 = 0.
However, only x = 8 will verify
x − 4 = √(2x).
A graphing calculator will show
that the area enclosed by y = x − 4
and y = √(2x) has a domain of 0 ≤ x ≤ 8.
Now build the Area Integral as ∫[from x=0 to x=8]{(2x)0.5 − (x − 4)}.
Translate this Integral to [(2√2/3)x1.5 − 0.5x2 + 4x|[from x=0 to x=8]].
Determine [(2√2/3)(8)1.5 − 0.5(8)2 + 4(8)] as (2√2/3)(8)1.5 or 64/3.
Determine [(2√2/3)(0)1.5 − 0.5(0)2 + 4(0)] as 0 − 0 + 0 or 0.
The shaded or enclosed area is then 64/3 − 0 or 64/3 square units
(21.333333333 square units).
