Find the area of the region bounded by the parabola y=4x^2, the tangent line to this parabola at (2, 16), and the x-axis.?

the tangent line is [ y - 16 ] = 16 [ x - 2 ]..thus integrate over y in [ 0 , 16] of [ line equation x = minus parabola equation x = ]

dy/dx = 8x
Slope of tangent = 16
y-16 = 16(x-2)
y = 16x-16
This line cuts the x axis at x = 1
Area under the tangent = (1/2)*1*16 = 8
A = ∫4x^2 dx from 0 to 2 - 8
A = 4x^3/3 from 0 to 2 - 8
A = 32/3 - 8 = 8/3