Please express the eqn of the parab by y = 2x^2. " ^ " indicates exponentiation.
Find the equation of the tangent line to y = 2x^2 at (1,2):
dy/dx = 4x; now let x = 1. Thus, the slope of the TL at (1,2) is 4(1) = 4, and the eqn of the TL is y-2 = 4(x-1), or y-2 = 4x -4, or y = 4x -2.
To find the area bounded by the parabola, the line y = 4x-2 and the x-axis, we need to use horizontal strips and integrate with respect to y instead of to x.
The smallest y value will be 0 (this is the x-axis), and the largest will be 2 (this comes from the given point, (1,2) ).
Solve y = 2x^2 for x: x^2 = y/2, so that x = sqrt(y/2) for y = 0 to y = 2.
y+2
Solve y = 4x - 2 for x: 4x = y + 2, so that x = -------
4
Thus, the length of each horiz. strip of width dy is given by
(y+2)
------- - sqrt(y/2), since the first term represents the larger x-value and the
4 second term represents the smaller x-value.
Then the area is the integral from y=0 to y=2 of
(y+2)
[ -------- - sqrt(y/2) ] dy
4
Can you do this integration? If not, ask specific questions so that I could help you further.