Find the area of the region bounded by the parabola y=x^3 , the tangent line to this parabola at (2,8), and the x-axis.

Answer 1

the function is y = x^3 because of (2,8)

y = x^3 ---> the inverse is y = x^(1/3)

y' = 3x^2

the slope at x=2 is M=3(2)^2 =12

B =y - mx = 8 - (12)(2) = 8 - 24 = -16

the tangent line is y = 12x-16

the inverse is y = (1/12)(x) + 4/3

integral:

((1/12)x - 4/3 ) - x^(1/3) , x = (0,8)

=

[(1/24) x^2 + (4/3) x - (3/4) x^(4/3)], x=(0,8)

x=0 the limit is zero

x=8: (1/24)(8^2) - (4/3)(8) - (3/4)(8)^(4/3) =

64/24 + 32/3 - (3/4)(16) =

64/24 + 32/3 - 12 =

8/3 +32/3 - 36/3 =

40/3 - 36/3 = 4/3